meromorphic functions

Lectures on Meromorphic Functions

By

W.K. Hayman

Notes by

K.N. Gowrisankaran

and

K. Muralidhara Rao

Contents

1 Basic Theory 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Poisson-Jensen Formula . . . . . . . . . . . 11.3 The Characteristic Function . . . . . . . . . . . 51.4 Some Inequalities . . . . . . . . . . . . . . . . . 111.5 . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 The Ahlfors-Shimizu Characteristic: . . . . . . . 151.7 Functions in the plane . . . . . . . . . . . . . . 231.8 Representation of a function.... . . . . . . . . . . 291.9 Convergence of Weierstrass products . . . . . . 33

2 Nevanlinna’s Second Fundamental Theorem 412.1 . . . . . . . . . . . . . . . . . . . . . . . . . . 412.2 Estimation of the error term . . . . . . . . . . . 442.3 . . . . . . . . . . . . . . . . . . . . . . . . . . 522.4 Applications . . . . . . . . . . . . . . . . . . . 542.5 Picard values of meromorphic... . . . . . . . . . 612.6 Elimination ofN(r, f ) . . . . . . . . . . . . . . 682.7 Consequences . . . . . . . . . . . . . . . . . . . 71

3 Univalent Functions 733.1 Schlicht functions . . . . . . . . . . . . . . . . 733.2 Asymptotic behaviour . . . . . . . . . . . . . . 95

iii

Part I

Basic Theory

1.11

We shall develop in this course Nevanlinna’s theory of meromorphicfunctions. This theory has proved a tool of unparallelled precision forthe study of the roots of equationsf (z) = a, f (1)(z) = b, etc. whethersingle or multiple and their relative frequency. Basic to this study is theNevanlinna CharacteristicT(r) which is an indication of the growth ofthe function f (z). We shall see in Theorem 2 that for every aT(r) is,apart from a bounded term, the sum of two componentsm(r, a)+N(r, a)of which the second measures the number of roots of the equation f (z) =a in |z| < r and the first the average closeness off (z) to a on |z| = r.The second fundamental theorem shows that in general the second termdominates and many applications giving well beyond Picard’s theoremresult.

1.2 The Poisson-Jensen Formula

We shall start with Poisson-Jensen formula which plays a fundamentalrole in our study.

Theorem 1. If f (z) is meromorphic in|z| ≤ R and has there zeros ap

and poles bν and if ζ = rei , f (ζ) , 0, then for0 ≤ r ≤ R we have

log f (reiθ) =12π

2π∫

0

log |(Reiφ)|(R2 − r2)dr

R2 − 2Rrcos (φ − θ) + r2+

∑

µ

log

∣

∣

∣

∣

∣

∣

R(ζ − aµ)

R2 − aµζ

∣

∣

∣

∣

∣

∣

1

2 Basic Theory

−∑

ν

log

∣

∣

∣

∣

∣

∣

R(ζ − bν)

R2 − bνζ

∣

∣

∣

∣

∣

∣

(1.1)

Proof. Let f (z) , 0, in |z| ≤ 1. Then, since we can define an analytic2

branch of logf (z) in |z| ≤ 1, we have by the residue theorem

12πi

∫

|z|=1

log f (z)dzz= log f (0)

By change of variable,

12π

2π∫

0

log f (eiφ)dφ = log f (0)

and now taking real part on both sides

12π

2π∫

0

log | f (eiφ)|dφ = log | f (0)|

For anyζ with |ζ | < 1, we effect the conformal transformationw =z− ζ

1− ζzfor the integral

∫

|z|=1

log f (z)dzz

. This in turn becomes,

12πi

∫

|w|=1

logϕ(w)dww= log f (ζ) where ϕ(w) = f {z(w)}

so thatϕ(0) = f (ζ). Substituting in the integralz = eiφ and taking realpart we get,

12π

2π∫

0

log | f (eiφ)|1− 2r cos(φ − θ) + r2

(1− r2) dφ = log | f (ζ)|, ζ = reiθ

�

Basic Theory 3

Now for the functionf (z) with polesbν and zerosaµ none of thembeing on|z| = 1, let us define

ψ(z) = f (z)

π(z− bν)

(1− bνz)

π(z− aµ)

(1− aµz)

On |z| = 1, |ψ(z)| = | f (z)| and the function has no zeros or poles in|z| ≤ 1. 3

By the above result,

12π

2π∫

0

log |ψ(eiθ)|(1− r2)

1− 2r cos(φ − θ) + r2dφ = log |ψ(ζ)|

ζ = reiθ , r < 1

Substitution forψ gives the theorem forR = 1. In the case when thereare poles and zeros on the circumference of the unit circle weproceedas follows. We have only to show that iff (z) has no zeros or poles in|z| < 1, but has poles and zeros on|z| = 1, then

12πi

∫

|z|=1

log f (z)dzz= log f (0)

For if f (z) has zeros and poles in|z| < 1 we can considerψ(z), in placeof f (z). Further we can assume that there is only one zero (the case ofpole being treated in the same manner) on|z| = 1. For the case whenf (z) has a finite number (it can have at most only a finite number) ofzeros (poles) can be treated similarly.

Let thereforez = a, |a| = 1 be a zero off (z) on |z| = 1. Let P bethe pointz = a and consider a circle of radiusρ < 1 aboutP, ρ beingsmall. Consider the contourS QR(fig.) Inside it f (z) has no zeros orpoles. Hence by the residue theorem,

∫

log f (z) dz = log f (0). Thus itis enough to prove that

∫

QR

log f (z) dz tends to zero asρ tends to zero.

4 Basic Theory

Let z = a be zero of orderk. Then f (z) = (z− a)kλ(z), λ(a) , 0, ina certain neighbourhood ofa; and we can assume the choice ofρ such4

that this expansion is valid within and on the circle of radius ρ aboutP.∫

QR

log f (z)dzz= k

∫

QR

log(z− a)dzz+

∫

logλ(z)dzz

Sinceλ(z) remains bounded the second integral tends to zero. So we

have only to prove that∫

QR

log(z− a)dzz

tends to zero asρ→ 0, Now,

∣

∣

∣

∣

∣

∣

∣

∣

∣

∫

QR

log(z− a)dzz

∣

∣

∣

∣

∣

∣

∣

∣

∣

≤ max

{∣

∣

∣

∣

∣

log |(z− a)||z|

∣

∣

∣

∣

∣

}

πρ ≤ [log(1/ρ) + 0(1)]πρ → 0 if ρ ≤12

This proves the result, in the case when the functionf (z) has zeros orpoles on the unit circle. In caseR , 1, we consider the functionf (Rz)instead of f (z) and arrive at the result. Hence the theorem is provedcompletely.

Corollary. In the special case whenζ = 0 we get the Jensen’s formula

(1.2) log| f (0)| =12π

2π∫

0

log | f (Reiφ)|dφ +∑

µ

log∣

∣

∣

∣

∣

aµR

∣

∣

∣

∣

∣

−∑

ν

log∣

∣

∣

∣

∣

bνR

∣

∣

∣

∣

∣

the summation ranging over poles and zeros of f(z) in |z| ≤ R. The aboveformula does not hold if zero is a pole or a zero of f(z). If f (0) = 0 or∞

6 Basic Theory

The first term is zero, in consequence of the factn(t) = 0, near zero.We write n(r, f ) for the number of poles off (z) in |z| ≤ r, so that

n(r, 1/ f ) is equal to the number of zeros off (z) in |z| ≤ r. We defineN(r, f ) to be

(3)

r∫

0

n(t, f )dtt

If f (0) = ∞ we defineN(r, f ) =r∫

0

[n(t, f ) − n(0, f )]dtt+ n(0, f ) log r.6

Then the formula (1.2) becomes, forf (0) , 0,∞

log | f (0)| =12π

2π∫

0

log+ | f (reiθ)|dθ −12π

2π∫

0

log+1

| f (reiθ)|dθ + N(r. f )

−N(r, 1/ f ).

We define,

(1.4) T(r, f ) = N(r, f ) +m(r, f )

where,

(1.5) m(r, f ) =12π

2π∫

0

log+ | f (reiθ)|dθ

Again (1.2) takes the form, forf (0) , 0,∞

(1.2′) T(r, f ) = T(r, 1/ f ) + log | f (0)|

If f (z) ∼ CλZλ near z = 0, whereλ , 0, then we obtainT(r, f ) = T(r, 1/ f ) + log |Cλ|. In future such modifications will be takenfor granted.

The functionT(r, f ) is called theCharacteristic Functionof f (z).This is the Nevanlinna characteristic function.

Theorem 2. First fundamental theorem.

Basic Theory 7

For any finite complexa,

T(r, f ) = T[r, 1/( f − a)] + log | f (0)− a| + ǫ(a)

where |ǫ(a)| ≤ log+ |a| + log 2.

Proof. We note that log+ |z1 + z2| ≤ log+ z1| + log+ |z2| + log 2

and log+ |z1 − z2| ≥ log+ |z1| − log+ |z2| − log 2. Whence, 7

log+ | f (z) − a| − log+ | f (z)| ≤ log 2+ log+ |a|

Integrating we get,

− log 2− log+ |a| +m(r, f − a) ≤ m(r, f ) ≤ log 2+ log+ |a| +m(r, f − a)

Since f and f − a have the same poles,

N(r, f ) = N(r, f − a).

Therefore,

T(r, f − a) − log+ |a| − log 2≤ T(r, f ) ≤ log 2+ log+ |a| + T(r, f − a)

That is,|T(r, f ) − T(r, f − a)| ≤ log 2+ log+ |a|

T(r, f ) + T(r, f − a) + ǫ(a), where |ǫ(a)| ≤ log 2+ log+ |a|

From (1.2′) we have,

T(r, f ) = T

(

r,1

f − a

)

+ log | f (0)− a| + ǫ(a)

where|ǫ(a)| ≤ log 2+ log+ |a|. Hence the theorem is proved. �

If we write m(r, a), N(r, a) for m

(

r,1

f − a

)

, N

(

r,1

f − a

)

, thenm(r, a)

represents the average degree of approximation off (z) to the value a onthe circle |z| = r and N(r, a) the term involving the numbers of zerosof f (z) − a. Their sum can be regarded as the total affinity of f (z) forthe valuea and we see than apart from a bounded term the total affinityfor every value ofa. However, the relative size of the two termsm, Nremains in doubt. We shall see in the second fundamental theorem thatin general it isN(r, a) that is the larger component. The value ofa forwhich this is not the case will be called exceptional.

8 Basic Theory

Let us now consider some examples8

1. Let f (z) = P(z)/Q(z), P(z) andQ(z) being polynomials of degreesm and n respectively, and prime to each other, that is have nocommon roots.

Then for larger, | f (z)| ∼ Crm−n.

So if m is greater thann, m(r, f ) = (m− n) log r + 0(1) and since

log+ x = 0, for 0 < x ≤ 1,m(r, 1/ f ) = 0

and m(r, 1/( f − a)) = 0.(1.6)

for large r and fixeda. Since f (z) = ∞ hasn roots in the openplane,n(t, f ) = n, for t > t0 andN(r, f ) = n log r + 0(1). AgainT(r, f ) = m(r, f ) + N(r, f ) which is equal tomlog r + 0(1) for afinite again by (1.6) and theorem 2,N(r, 1/( f−a)) = mlog r+0(1),

a finite. Thus

[

N

(

r,1

f − a

)

/T(r, f )

]

→ 1 asr → ∞. In this case

a = ∞ is the only exceptional value.

Similar conclusions follow ifm is less thann by taking 1/ f inplace of f in the above discussion. In this casea = 0 = f (∞) isthe only exceptional value.

If m is equal ton, f = c+0(z−λ) (say). Or writing,f − c ∼ b(z−λ),we see thatm(r, 1/( f − c)) = λ log r + 0(1)

m(r, 1/( f − a)) = 0(1), when a , c

m(r, f ) = 0(1),N(r, f ) = n log r + 0(1)

ThusT(r, f ) = n log r + 0(1) and,

N(r, a) = n log r + 0(1), a , c

N(r, c) = (n− λ) log r + 0(1).

Thus in any case

T(r, f ) = (Max . m, n) log r + 0(1)

i.e., T(r, f ) ∼ (Max . m, n) log r.

Basic Theory 9

Thus in the case of a rational function there is always only one9

exceptional value, viz.,f (∞)

2. Let f (z) = ez. In this case the value ofm(r, f ) is,

m(r, f ) = 1/2π

2π∫

0

log+ er cosθdθ

= 1/2π

π/2∫

−π/2

cosθ dθ

= r/π

because cosθ ≤ 0 in π/2 ≤ θ ≤ π, −3π/2 ≤ θ ≤ −π/2 and solog+ er cosθ is equal to zero.

Sinceez has no poles,N(r, f ) = 0. ConsequentlyT(r, f ) is equal tom(r, f ) which in turn is equal tor/π.

We employ the notationm(r, a) = m(r, 1/( f − a)) for finite a andm(r,∞) = m(r, f ), and similarly with the functionsn, N, andT. Wehave|ez − a| ≥ ||ez| − |a|| = |er cosθ − |a|| if z = reiθ. If a , 0, we havefor large r er < |a| < er . Therefore|a| = er cosα for 0 ≤ α ≤ π. Thus

2m(r, a) =2π∫

0

log+1

|ez − a|dθ

≤

2∫

0

log+1

|er cosθ − er cosα|dθ

= 2π log+1

er cosα + 2

2∫

0

log+1

|er(cosθ−cosα) − 1|dθ

If cosθ − cosα ≥ 0, we have|er(cosθ−cosα) − 1| 10

= r(cosθ − cosα) +r2

2(cosθ − cosα)2

+ · · · ≥ r(cosθ − cosα)

Basic Theory 11

Hence

T(r, a) = T(r, f ) + 0(1)= r/π + 0(1).

i.e. N(r, a) +m(v, a) = r/π + 0(1).

i.e. N(r, a) = r/π + 0(1).

This shows thatN(r, a)T(r)

→ 1.

1.4 Some Inequalities

We have already seen that

log+ z1 + z2 log+ z1 + log+ |z2| + log 2

More generally,

log+∣

∣

∣

∣

∣

∣

∣

ν=n∑

ν=1

zν

∣

∣

∣

∣

∣

∣

∣

≤ log+ |nmax|zν|∣

∣

∣

∣

≤ log+ n+ log+ |max|zν|∣

∣

∣

∣

≤ log+ n+ν=n∑

ν=1

log+ |zν|

Hence,

m

r,N

∑

n=1

fn

≤

n=N∑

n=1

m(r, fn) + logN

Now F =N∑

n=1fn(z) has poles only where thefn(z) have poles and

the multiplicity of such pole is at most the maximal multiplicity of thepoles of fn(z) which is not greater than the sum of the multiplicities.This gives

N(r, F) ≤N

∑

n=1

N(r, fn)

and T

r,N

∑

n=1

fn

≤

N∑

n=1

T(r, fn) + log N

Basic Theory 13

1.513

Theorem 3 (H. Cartan). We have T(r, f ) = (1/2π)2π∫

0

N(r, eiθ)dθ +

log+ | f (0)| that is, T(r, f ) apart from a constant term is the average ofN(r, a), for a on the circle|a| = 1.

For the proof we require the following lemma.

Lemma 1. If a is complex,

1/2π

2π∫

0

log |a− eiθ |dθ = log+ |a|

This can be proved in various ways. If|a| ≥ 1, z− a has no zeroes in|z| < 1, and apply Poisson formula withf (a) = z− a, R= 1.

Then

12π

2π∫

0

log |eiθ − a|dθ = log |a| = log+ |a|, |a| ≥ 1

If

|a| < 1 write, |a− eiθ | = |a||1− eiθ/a| = |a||1− e−iθ/a|

= |a||eiθ − i/a|

and we get by the first part, since∣

∣

∣

∣

∣

1a

∣

∣

∣

∣

∣

> 1,

12π

2π∫

0

log |eiθ − a|dθ = log |a| + log∣

∣

∣

∣

∣

1a

∣

∣

∣

∣

∣

= 0 = log+ |a|.

To prove the theorem consider now Jensen’s formula applied to f (z)−eiθ,

log | f (0)− eiθ | =12π

2π∫

0

log | f (reiφ) − eiθ |dφ + N(r, f ) − N(r, 1/( f − eiθ))

14 Basic Theory

sinceN(r, f ) = N(r, f − eiθ) ( f and f − eiθ have the same poles). Nowintegrate both sides with respect toθ and invert the order of integrationin the first term in the right hand side which we can do by Fubini’sTheorem, because the integrand is bounded above by log+[| f (reiφ)| + 1]14

and this is positive, integrable and independent ofθ.

12π

2π∫

0

log | f (0)− eiθ |dθ =12π

2π∫

0

dθ12π

2π∫

0

log | f (reiφ) − eiθ |dφ

+12π

2π∫

0

N(r, f )dθ −12π

2π∫

0

N(r, eiθ)dθ

and by the previous lemma,

log+ | f (0)| =12π

2π∫

0

log+ | f (eiφ)|dφ + N(r, f ) −12π

2π∫

0

N(r, eiθ)dθ

= T(r, f ) −12π

2π∫

0

N(r, eiθ)dθ

which gives the result.

Corollary 1. T(r, f ) is an increasing convex function oflog r. It is alsoeasily seen that m(r, f ) need not be either an increasing or a convexfunction oflog r (e.g. by considering rational functions). In fact fromtheorem 3,

d(T(r))d log r

=d

d log r12π

2π∫

0

dθ

r∫

0

n(t, eiθ)dtt=

12π

2π∫

0

n(r, eiθ)dθ

and the right hand side is non-negative and non-decreasing with r.

Corollary 2.12π

2π∫

0

m(r, eiθ)dθ ≤ log 2.

Basic Theory 15

In fact by theorem 2,

m(r, eiθ) + N(r, eiθ) = T

(

r,1

f − eiθ

)

= T(r, f ) − log | f (0)− eiθ)|+ < log 2>

where< ǫ > denotes any quantity that is less thanǫ in modulus. 15

Integrating,

12π

2π∫

0

m(r, eiθ)dθ +12π

2π∫

0

N(r, eiθ)dθ

= T(r, f ) − log+ | f (0)|+ < log 2>

by the lemma.So

12π

2π∫

0

m(r, eiθ)dθ + T(r, f ) − log+ | f (0)|

= T(r, f )+ < log 2> − log+ | f (0)|

which is the required result. i.e.

12π

2π∫

0

m(r, eiθ)dθ =< log 2>

1.6 The Ahlfors-Shimizu Characteristic:

We have defined the Nevalinna characteristic function of a meromorphicfunction f (z). We now proceed to define the characteristic function afterAhlfors [1] and Shimizu [1]. Prior to that let us prove the followinglemma.

Lemma 2 (Spencer). Suppose that D is a bounded domain in the com-plex plane, whose boundary is composed of a finite number of analytic

16 Basic Theory

curves,γ, and that G(R) is twice continuously differentiable on the setof values R assumed by| f (z)| in D and onγ, f (z) being regular in D andonγ. Then

∫

γ

∂

∂nG(| f (z)|)ds=

"

D

g(| f (z)|)| f ′(z)|2dx dy

where g(R) = G′′(R)+(1/R)G′(R) and we differentiate out of the domain16

along the normal in∂

∂n.

Let us make use of Green’s formula which under the hypothesisofthe lemma gives,

∫

γ

∂

∂nG(| f |)ds=

"

D

∇2[G(| f |)]dx dy∇2=

∂2

∂x2+∂2

∂y2

Suppose first thatf (z) has no zeros forz in D. Putu = log | f |, in orderto facilitate the easy calculation of∇2G(| f |). u is a harmonic function.Therefore,∇2u = 0. Now | f | = eu, G(| f |) = G(eu). Calculating thepartial derivatives with respect tox andy we see that,

∂2

∂x2[G(eu)] =

(

∂u∂x

)2{

euG′(eu) + e2uG′′(eu)}

+ euG′(eu)∂2u

∂x2

and

∂2

∂y2

(

G(eu))

=

(

∂u∂y

)2{

euG′(eu) + e2uG′′(eu)}

+G′(eu)eu∂2u

∂y2

And u = Rl. log f (z) and∂u∂y− i

∂u∂y=

ddz

(log f ) = f ′/ f .

Hence on addition we get,

∇2[G(eu)] =| f ′(z)|2

| f (z)|2eu {

G′(eu) + euG′′(eu)}

since∇2u = 0.

Basic Theory 17

Writing | f (z)| = R,

∇2[G(R)] = | f ′(z)|2[G′(R) + RG′′(R)]/R

= g(R)| f ′(z)|2

whereg(R) = (1/R)G′(R) +G′′(R). 17

This establishes the result provided thatf (z) , 0 for z in D. Other-wise, exclude the zeros (which are necessarily finite in number) of f (z)

in D, by circles of small radii, and in these∂

∂nG(| f |) is bounded, since

by hypothesisG(R) is continuously differentiable nearR= 0. Hence thecontribution to the left hand side of the formula tends to zero with theradii of the circles, and so the lemma is proved.

Let us now specialise withG(R) = 12 log(1+ R2) and f (z) a mero-

morphic function in|z|, r with no poles on|z| = r. Exclude the poles in|z| < r by circles of radiiρ (small). Let us apply the lemma to the regionD consisting of|z| < r with the poles excluded. We find in this case,

g(R) = (1/R)G′(R) +G′′(R) = 2/(1+ R2)2, and∫

γ

∂

∂nG(| f |dx=

"

D

g(| f |)| f ′(z)|2dx dy.

γ consists of the circumference of|z| = r and the smaller circles. Near apolez0 of orderk, of f (z)

| f (z)| ∼ |0|/λk |z− z0| = λ

and

G(| f |) =12

log(1+ | f |2) = log | f | + 0(1).

that is,G(R) = k logγλ + 0(1)

Therefore,∂

∂n[G(R)] = −

∂

∂ρ

(

k log1ρ

)

+ 0(1)

18 Basic Theory

since in the lemma the derivative∂

∂nis along the normal directed out of

the region and hence here into the circle (isolatingz0). 18

∂

∂n[G(R)] = k/ρ + 0(1).

Hence the contribution to the left hand side in the lemma is 2πρ[k/ρ +0(1)], that is, 2πk + 0(ρ). Adding over all the circles and lettingρ tendto zero,

2πn(r, f ) +∫

|z|=r

∂

∂n[log(1+ | f |2)

12 ]ds

=

"

|z|<r

2| f ′(z)|2

[1 + | f (z)|2]2dx dy

that is again,

n(r, f ) +12π

2π∫

0

∂

∂r

[

log√

1+ | f (reiθ)|2]

r dθ

=1π

"

|z|<r

| f ′(z)|2

[1 + f (z)2]2dx dy.

We call the right hand sideA(r).Integrate both sides from 0 to ‘r ’ with respect tor after dividing by

r throughout to get,

N(r, f ) +12π

2π∫

0

log(1+ | f (rei )|2)12 dθ − log(1+ | f (0)|2)

12

=

r∫

0

A(t)dtt

.(1.7)

The integration is justified since both sides are continuousand haveequal derivatives except for the isolated values ofr for which |z| = r

Basic Theory 19

contains poles of the functionf (z). Now log+ R ≤ log(1 + R2)12 ≤

logR+ 12 log 2, if R≥ 1 so that log(1+R2)

12 ≤ log+R+ < 1

2 log 2> and19

if R ≤ 1, 0 ≤ log(1+ R2)12 ≤ 1

2 log 2 = 12 log 2+ log+R so that in all

cases log(1+ R2)12 ≤ log+ R+ < 1

2 log 2>. Hence log[1+ | f (reiθ)|2]12 =

< 12 log 2> + log+ | f (reiθ)| substituting,

r∫

0

A(t)dtt=

12π

2π∫

0

log+( f (reiθ)|dθ + N(r, f )+ <12

log 2> − log+ | f (0)|

= m(r, f ) + N(r, f )+ <12

log 2> − log+ | f (0)|

= T(r, f )+ <12

log 2> − log+ | f (0)|

Now if we put,

T0(r, f ) =

r∫

0

A(t)dtt

We get

T0(r, f ) = T(r, f )+ <12

log 2> − log+ | f (0)|

Definition. T0(r, f ) is called Ahlfors-Shimizu Characteristic function off (z).

Interpretation of (1.7)

20 Basic Theory

Consider the Riemann Sphere of diameter one lying over thew-planeand touching it atw = 0. To every pointw corresponds the pointP(w)in which the lineNw (fig) intersects the sphere. From the figureNP =cosθ = 1/(1+ R2)

12 , R= |w|. NP is called the chordal distance between

w and∞ and is denoted by [w,∞].Then20

12π

2π∫

0

log[1+ | f (reiθ)|2]12 dθ =

12π

2π∫

0

log1

[ f (z),∞]dθ

and so is the average of the chordal distances betweenf (z) and infinity,whenz runs over|z| = r. The left hand side of the above equation is analternative form(r, f ) in the sense that it differs fromm(r, f ) by less than12 log 2.

It is easy to see that ifds is an element of length in the planeanddσ the corresponding element on the Riemann-sphere thendσ =ds/(1 + R2). Hence ifdu dv is an element of area in thew-plane, thenthe corresponding element of area on the sphere isdu dv/(1 + R2)2. Ifdx dyis an element of area inz-circle, |z| < r, its image in thew-plane is| f ′(z)|2dx dy, since the element of length is multiplied by| f ′(z)|. Thusthe element of area corresponding todx dyon the Riemann-w-sphereis precisely| f ′(z)|2dx dy/[1 + | f (z)|2]2. Therefore, we interpretπA(r)as the area on the Riemann-w-sphere of the image, with due count ofmultiplicity (the mapw = f (z) may not be one-one, and more than onelement of arcx in thez-plane may go into the same element of arcx inthew-plane) of|z| < r by w = f (z). Since the area of the sphere isπ, A(r)itself being the area of the image divided by the area of the sphere maybe interpreted as the average number of roots in|z| < r of the equationf (z) = w, asw moves over the Riemann sphere.

21

We have seen above thatA(r) =1π

area of image of|z| < r by f (z) on the

Riemann Sphere. So thatA(r) can be interpreted as the average valueof the numbern(r, a) of roots of f (z) = a, in |z| < r asa runs over thecomplex plane.

Basic Theory 21

Let us now rotate the sphere. This corresponds to a one-one confor-mal map of the closed plane onto itself and in fact to a bilinear map ofthe typew′ = eiλ(1+aw)/(w−a) (for the proof refer Caratheodery, [1]).Further under this rotationA(r) will remain unaltered.

Setϕ(z) = eiλ[1 + a f(z)]/[ f (z) − a], and observe that [ϕ(z),∞] =[ f (z), a], wherew, a is the chordal distance between the points corre-sponding tow and a on the Riemann Sphere. In particular [w,∞] ={1+ |w|2}

12 . If we apply our previous result toϕ(z), we get

r∫

0

A(t)dtt= N(r, ϕ) +

12π

2π∫

0

log1

[ϕ(reiθ),∞]dθ − log

1[ϕ(0),∞]

= N(r, a) +12π

2π∫

0

log1

[ f (reiθ), a]dθ − log

1[ f (0), a]

.

Thus we have proved the theorem.

Theorem.For every complexa anda = ∞

r∫

0

A(t)dtt= N(r, a) +

12π

2π∫

0

log1

[ f (reiθ), a]dθ − log

1[ f (0), a]

.

The new chordal distance is 22

[w, a] =1

1+∣

∣

∣

∣

∣

1+ aww− a

∣

∣

∣

∣

∣

2

12

=|w− a|

[(1 + |a|2)(1+ |w|2)]12

Thus in this above theorem, the expressionm0(r, a) replacesm(r, a) ofthe Theorem 2, where,

m0(r, a) =12π

2π∫

0

log[(1+ |a|2)(1+ | f (reiθ)|2)]12

| f (reiθ) − a|dθ

Unlike theorem 2 theorem 2’ is exact. Notice thatm0(r, a) is always non-negative because the chordal distance between any two points is alwaysis less or equal to one.

22 Basic Theory

Corollary. If f (z) is meromorphic in the plane then

r∫

1

n(t, a)dtt≤

r∫

1

A(t)dtt+C.

for all a where C is independent of a and r(> 1) but may depend on f .

Proof.

r∫

0

A(t)dtt= N(r, a) +m0(r, a) −m0(0, a)

1∫

0

A(t)dtt= N(1, a) +m0(1, a) −m0(0, a)

subtracting, sinceN(r, a) =r∫

0

n(t, a)dtt

,

r∫

1

n(t, a)dtt=

r∫

1

A(t)dtt+m0(1, a) −m0(r, a)

≤

r∫

1

A(t)dtt+max

am0(1, a)

sincem0(r, a) is greater or equal to zero.23

The maximum on the right is finite. In fact, for variablea and fixedr m0(r, a) is continuous inaand is bounded asamoves over the Riemannsphere. This is evident if the functionf (z) , a on |z| = r. If f (z) = a, ata finite number of points the argument is similar to that in thePoisson-Jensen formula. PuttingC = Maxa m0(1, a) the result is obtained. �

Remark. By an inequality for real positive functions due to W.K. Hay-man and F.M. Stewart [1] and using corollary 1 we deduce that if f (z) ismeromorphic and not constant in the plane and

n(r) = supa·n(r, a) and ǫ > 0,

Basic Theory 23

then

(1.8) n(r) < (e+ ǫ)A(r) . . . . . . . . .

on a setE of r having the following property. IfE(r) is the part ofE in

[1, r] then∫

E(r)

dtt≥ c(ǫ) log r for all larger, wherec(ǫ) depends only on

ǫ.

The following are some open problems:

(a) Cane be replaced by a smaller constant and in particular by onein (1.8)?

(b) Does (1.8) hold for all larger; or for all larger except a small set?24

(c) Can we assert

r∫

1

n(t)dtt< (constant).A(r)

for some arbitrarily larger?

1.7 Functions in the plane25

Let S(r) be a real function≥ 0, and increasing for 0≤ r ≤ ∞. Theorderk and thelower orderλ of the functionS(r) are defined as

kλ

}

= limr→∞

[log S(r)]/ log r.

The order and the lower orders of the function always satisfythe rela-tion, 0≤ λ ≤ k ≤ ∞.

If 0 < k < ∞, we distinguish the following possibilities,

(a) S(r) is maximal type if

C = limS(R)/Rk= infinity.

24 Basic Theory

(b) S(r) is of mean type of orderk if 0 < C < ∞.

(c) S(r) is of minimal type ifC = 0.

(d) S(r) is of convergence class if,

∞∫

1

S(t)dt/tk+1 converges.

Note that ifS(r) is of orderk andǫ > 0,

S(r) < rk+ǫ for all large r.

andS(r) > rk−ǫ for some larger.

(This follows from the definition of order ofS(r).)It can be seen that ifS(r) is of orderk and of convergence type

i.e., (d) then it is of minimal type, (c). In fact in this case

∞∫

r0

S(r)

rk+1dr < ǫ if r0 > t(ǫ)

Then262r0∫

r0

S(r)

rk+1dr < ǫ

and sinceS(r) increases withr,

2r0∫

r0

S(r)

rk+1dr ≥

S(r0)

(2r0)k+1r0

andS(r0)2−(k+1)/r0k < ǫ for r0 > t(ǫ)

that is,limr→∞

S(r)/rk= 0.

Basic Theory 25

So if (d) holdsS(r) is of minimal type. We also note that ifk′ is greaterthan the order ofS(r) then,

∞∫

1

S(r)dr/rk′+1 converges.

and ifk′ is less thank∞∫

1

S(r)

rk′+1dr diverges.

We have next

Theorem 4. If f (z) is a regular function for|z| ≤ R and

M(r, f ) = Max|z|=r f (z) then,

T(r, f ) ≤ log+ M(r, f ) ≤r + RR− r

T(R, f ) for 0 < r < R.

Proof. Since f (z) is regular in|z| ≤ Randr < Rwe have,

T(r, f ) = T(r) = m(r, f ) =12π

2π∫

0

log+ | f (reiθ)|dθ

≤12π

Max|z|=r log+ | f (reiθ)|2π

= log+ M(r, f )

To prove the other side of the above inequality of the theoremwe dis- 27

tinguish between the two cases (i)M(r) < 1 or (ii) M(r) ≥ 1. In the case(i), log+ M(r) = 0 andT(r, f ) × (R+ r)/(R− r) being non-negative theinequality holds good. Hence we can now suppose thatM(r) ≥ 1, andin which case log+ M(r) = log M(r).

The Poisson-Jensen formula gives, forz= reiθ

log | f (z)| =12π

2π∫

0

log | f (Reiφ)|(R2 − r2)

R2 + 2Rrcos(φ − θ) + r2dφ.

26 Basic Theory

−∑

log+∣

∣

∣

∣

∣

∣

(R2 − aµz)

R(z− aµ)

∣

∣

∣

∣

∣

∣

in the above equalityaµ runs through all the zeros off (z). For, as regardsthe zeros with modulus less or equal toR

log+∣

∣

∣

∣

∣

∣

(R2 − aµz)

R(z− aµ)

∣

∣

∣

∣

∣

∣

= log

∣

∣

∣

∣

∣

∣

(R2 − aµz)

R(z− aµ)

∣

∣

∣

∣

∣

∣

since

∣

∣

∣

∣

∣

∣

(R2 − aµz)

(z− aµ)R

∣

∣

∣

∣

∣

∣

≥ 1.

and for the zerosaµ with modulus greater thanR,

log+∣

∣

∣

∣

∣

∣

(R2 − aµz)

R(z− aµ)

∣

∣

∣

∣

∣

∣

= 0

and Poisson’s formula is unaffected. �

Clearly,

log | f |(R2 − r2)

R2 − 2Rrcos(φ − θ) + r2≤ log+ | f |

(R2 − r2)R2 − 2Rrcos(φ − θ) + r2

≤R+ rR− r

log+ f −R+ rR− r

log+ f

28

log | f (z)| ≤12π

2π∫

0

(R2 − r2)(r2 + R2 − 2Rr)

log+ | f (Reiφ)|dφ

= (1/2π)

2π∫

0

(R+ r/(R− r)) log+ | f (Reiφ)|dφ

log | f (z)| ≤ (R+ r/(R− r))T(R, f ).

This being true for allz, |z| = r, holds good for thezat which f (z) takesthe maximumM(r, f ).

Hence,log M(r, f ) ≤ (R+ r/R− r)T(R, f ).

Basic Theory 27

Corollary. If f (z) is an integral function, the order and type (a, b, c ord defined in the last article) of T(r) and log+ M(r) are the same.

From the left hand side of the inequality of Theorem 4, it followsthat the order and type ofT(r, f ) are not bigger than those of log+ M(r).In order to prove the reverse, let us substituteR= 2r in the right side ofthe theorem. We get,

log+ M(r, f ) ≤ 3T(2r, f )

Supposek is the order ofT(r, f ), then givenǫ > 0, T(r, f ) < ǫrk+ǫ for allr > r0 by the definition of order. Hence combining the two inequalities,

log+ M(r, f ) ≤ 3T(2r, f ) < 3ǫ(2r)k+ǫ for r > r0.

This implies that the order of log+ M(r, f ) is less or equal tok. If T(r)has mean type, we may takeǫ = 0. If T(r) has minimal type, we may29

in addition takec arbitrarily small. HenceT(r, f ) and log+ M(r, f ) bothhave same order and belong to the same order typea, b or c. In order tocomplete the corollary, we have to consider the case whenT(r, f ) is of(d) i.e., convergence type. Consider,

∞∫

r0

log+ M(r)

rk+1dr

∞∫

r0

log+ M(r)

rk+1dr ≤

∞∫

r0

3T(2r)

rk+1dr

k being the order ofT(r, f ).

=

∞∫

2r0

3.2k T(r)dr

rk+1

by change of variable.SinceT(r) is of convergence type, it follows from the above inequal-

ity that log+ M(r) also belongs to the same class. It is clear that thisrelation holds good in the reverse direction.

28 Basic Theory

We now proceed to define the order and the order type of a functionf (z) meromorphic in the plane on the above analogy.

Definition . Theorderand type of a function f(z) meromorphicin theplane aredefinedas theorderand type ofT(r, f ).

We note that this coincides with that of order and type oflog+ M(r, f ) if f (z) is an entire function.

Example. (1) If the function f (z) = ez, log M(r) = r andT(r) = r/π.30

The function has order one, mean type. In this case log+ M(r) andT(r) are of the same type, viz., mean type, though the values ofthe type are different. More precisely, forT(r), lim

r→∞T(r)/r = 1/π

(the order being 1) and for log+ M(r), limr→∞

(1/r) log+ M(r) = 1.

In the above example the ratio of the two super. limites that is,

limr→∞

T(r)

rk

limr→∞

log+ M(r)

rk

= 1

In general case for a function of orderk and mean type this ratio isbounded by 1. But it is still an open problem to find the best pos-sible lower bound. It can be shown easily from theorem 4, takingR = r[1 + (1/k)], that a lower bound is 1/e(2k + 1). In this connectionP.B. Kennedy [1] has given a counter example of a function of order kmean type in|argz| < π/2k, and bounded forπ/2k ≤ argz≤ π, providedk > 1

2, such that

log+ | f (reiθ)| ∼ rk coskθ for |θ| < π/2k

log+ | f (reiθ)| = 0(1)π

2k≤ |θ| < π

HenceT(r, f ) ∼ rk/k, log M(r) ∼ rk. Thuse(2k + 1) cannot be replacedby any constant less thankπ if k > 1

2.

Basic Theory 29

Remark . For functions of infinite orderT(r) and log+ M(r) no longerhave necessarily the same order of magnitude. Actually it has beenshown by W.K. Hayman and F.M. Stewart [1] that ifǫ > 0, logM(r) <(2e+ǫ)[T(r)+A(r)] for some larger. As an example considerf (z) = eez

31

log M(r) = er

and

T(r) ∼ er (2π3)12/r

12 .

1.8 Representation of a function in terms of its zeros and poles32

Definition. For any complex number z, and any integer q> 0,

E(z, q) = (1− z)ez+ 12z2+···+ 1

qzq.

Theorem 5(Nevanlinna). If f (z) is meromorphic in the plane with zerosaµ and poles bν and f(z) being of order at most q of the minimal typethen,

f (z) = zpePq−1(z) limR→∞

∏

1aµ |<RE

(

zaµ, q− 1

)

∏

1bµ |<RE

(

zbν, q− 1

)

where p is a suitable integer and Pq−1(z) is a polynomial of degree atmost q− 1.

Note that the theorem only asserts that the limit on the rightexists,but it does not indicate whether the products are convergentor divergent.

In fact if f (z) only satisfies the weaker condition that lower limit asr tends to infinity ofT(r)/rq is equal to zero (instead of the conditionassumed in the theorem viz., upper limit of the same is zero.), the resultstill holds provided thatR is allowed to tend to infinity through a suitablesequence of values, instead of all values.

To start with let us assume that the functionf (z) has no zero or pole33

30 Basic Theory

at zero. The Poisson-Jensen formula gives,

log | f (z)| =12π

2π∫

0

(R2 − r2) log | f (Reiφ)|R2 − 2Rrcos(φ − θ) + r2

dφ +∑

|aµ |<R

log

∣

∣

∣

∣

∣

∣

R(z− aµ)

(R2 − aµz)

∣

∣

∣

∣

∣

∣

z= reiθ+

∑

|bν |<R

log

∣

∣

∣

∣

∣

∣

R2 − bνzR(z− bν)

∣

∣

∣

∣

∣

∣

Both the sides are equal harmonic functionsv(z) (say) ofznear the point

z= reiθ where f (reiθ) , 0,∞. Let us operate∂v∂x− i∂v∂y

on both the sides.

We assume thatR is such thatf (z) , 0,∞ on |z| = R. Differentiatingunder the integral sign and observing that

Real

(

Reiφ+ z

Reiφ − z

)

=(R2 − r2)

R2 − 2Rrcos(φ − θ) + r2

We deduce,

f ′(z)f (z)

=12π

2π∫

0

log | f (Reiφ)|2Reiφ

(Reiφ − z)dφ

−∑

|aµ |<R

[

1(aµ − z)

−aµ

(R2 − aµz)

]

+

∑

|bν |<R

1(bν − z)

−bν

(R2 − bνz)

provided that there are no zeros or poles on|z| = R. Differentiating this34

q− 1 times,

(

ddz

)q−1 (

f ′(z)f (z)

)

=1π

2π∫

0

log | f (Reiφ)|Reiφ

(Reiφ − z)q+1

+ (q− 1)∑

|bν |<R

1(bν − z)q −

bqν

(R2 − bνz)q

Basic Theory 31

+ (q− 1)∑

|aµ |<R

1(aµ − z)q −

aqµ

(R2 − zaµ)q

Suppose now thatT(2r)/(r)q tends to zero asr tends to∞ either throughall values or through a suitable sequence of values, sayRk (which tendsto∞with k). Such a sequence of values exists by our assumptions. Alsoby decreasingRk slightly if necessary we may assume thatf (z) , 0,∞onz= Rk, we takeR= Rk in (1).

Then sincem(r, f ) ≥ 0,

T(2Rk, f ) ≥ N(2Rk, f ) ≥

2Rk∫

Rk

n(t, f )dtt≥ n(Rk, f ) log 2.

Thusn(Rk, f )Rqk tends to zero ask tends to∞. Similarly, n(Rk, 1/ f )/Rq

k→ 0, ask→ ∞) since,T(Rk, 1/ f ) = T(Rk, f )+ 0(1) andT(Rk, f )/Rq

k →

0. Our aim now is to show that some of the terms on the right of theequation 11.9 including the integral tend to zero uniformlyfor z on anybounded set ask tends to infinity. Then lettingk tend to infinity, we geta modified equation integrating whichq times we will get the result.

Now suppose that|z| <12

Rk. Then,|bνz| <12

Rk · Rk for |bν| < Rk. 35

and

|R2k − bνz| ≥ R2

k − |bνz| ≥12

R2k

Hence,∣

∣

∣

∣

∣

∣

∣

bqν

(R2k − bνz)

∣

∣

∣

∣

∣

∣

∣

<Rq

k

(12

R2k)q=

2q

Rqk

this inequality being true for all polesbν with |bν| < Rk. Therefore,summing up for all polesbν, |bν| < Rk

∣

∣

∣

∣

∣

∣

∣

∑ bqν

(R2k − bνz)

∣

∣

∣

∣

∣

∣

∣

≤∑

∣

∣

∣

∣

∣

∣

∣

bqν

(R2k − bνz)q

∣

∣

∣

∣

∣

∣

∣

<n(Rk, f )2q

Rqk

and hence the right hand side tends to zero uniformly ask tends to in-finity for z in any bounded set. A similar result holds good in the caseof the zeros of the function.

32 Basic Theory

Now coming to the integral,

|Rkeiφ − z)| ≥

12

Rk for |z| <12

Rk.

Hence the modulus of the integral on the right of (1.9) is at most,

(q!)Rk

π

2q+1

Rq+1k

2π∫

0

∣

∣

∣log | f (Rkeiφ)|

∣

∣

∣dφ

(q!)π

2q+1

Rqk

2π∫

0

log+ | f (Reiφ)|dφ +

2π∫

0

log+1

| f (Reiφ)|dφ

< (q!)2q+2

Rqk

[m(Rk, f ) +m(Rk, 1/ f )]

< (Const.)T(Rk, f )

Rqk

< (Const.)T(2Rk, f )

Rqk

→ 0

as k→ ∞.

Now the equation (1.9) takes the form,36

(

ddz

)q−1 f ′(z)f (z)

= limk→∞

Sk(z)

Where

Sk(z) = (q− 1)!

∑

|bν |<Rk

1(bν − z)

q−∑

|aµ |<Rk

1(aµ − z)

q

the convergence being uniform for any bounded set of values of z notcontaining any of the zeros or poles off (z).

By the uniform convergence we may therefore, integrate bothsides(q− 1) times along a suitable path from 0 toz to get,

f ′(z)f (z)

= limk→∞

∑

|bν |<Rk

1(bν − z)

−1bν−

z

b2ν

− · · · −zq−2

bq−1ν

Basic Theory 33

−∑

|aµ |<Rk

1(aµ − z)

−1aµ− · · · −

zq−2

aq−1µ

+ pq−2(z)

wherePq−2(z) is a polynomial of degree at mostq − 2. Now integrateboth the sides once more from 0 toz, and take exponentials,

f (z) = Pq−1(z) limk→∞

∏

|aµ |<Rk

(

1−zaµ

)

ez

aµ+

z2

2a2µ+···+ zq−1

(q−1)aq−1µ

∏

|bν |<Rk

(

1−zbν

)

ez

bν+··· 1

q−1

(

zbν

)q−1

Hence the result in the case whenf (0) , 0,∞. In case zero is a pole or37

zero of the function of orderp, consider the functionf (z)/zp and applythe result just obtained to get the theorem in its final form.

1.9 Convergence of Weierstrass products38

Let a1, a2, . . .an . . . be a sequence of complex numbers (none 0) withmoduli r1, r2, . . . , rn, . . . in the increasing order of magnitude. Letn(r)be defined as

n(r) = Sup.rk<r

k.

Then follows the result:

Lemma. If N(r) =r∫

0

n(t)t

dt then N(r) and n(r) have the same order and

type; and for any k, such that0 < k < ∞, the series∑

1/rkn and the

integrals∞∫

0

n(r)

rk+1dr, and

∞∫

0

N(r)dr

rk+1converge or diverge together.

Proof. By Riemann-Stieltjes’ integrals,

∑

rn<R

1/rkn =

R∫

0

1

tkdn(t)

34 Basic Theory

= k

R∫

0

n(t)dt

tk+1+ n(R)/Rk becausen(t) = 0

near 0. Suppose nowI1 =

∞∫

0

n(R)dR

Rk+1is less than infinity.n(R) has at

most convergence type of orderk. (Hence it is also of minimal type).Therefore upper limit ofn(r)/rk asr tends to infinity= zero, and

∑

1/rkn

converges tok(I1). The convergence of the series implies the conver-gence of the integral, becausen(R)/Rk ≥ 0. �

Now consider the integral,39

R∫

0

N(r)dr

rk+1= N(R)/(−k)Rk

+

R∫

0

dN(r)/krk

= −N(R)/kRk+

R∫

0

n(r)k

dr/rk+1

From this inequality we get that the integrals∞∫

0

N(r)dr/rk+1 and

∞∫

0

n(r)dr/rk+1 will converge or diverge together, due to the following

reason. The convergence of∞∫

0

N(r)dr

rk+1implies that lim

R→∞N(R)/Rk

= 0,

hence the convergence of∞∫

0

n(r)dr

rk+1follows. On the other hand if

∞∫

0

n(r)dr/rk+1 convergesR∫

0

N(r)dr

rk+1at once by comparison.

Therefore it remains to be proved thatn(r) andN(r) have the sameorder and type. Supposen(r) has orderk, then givenδ > 0, n(r) < crk+δ

for r greater thanr0

N(r) =

r∫

0

n(t)dtt

Basic Theory 35

=

r0∫

0

n(t)dtt+

r∫

r0

n(t)dtt

for r > r0

N(r) ≤

r0∫

0

n(t)dtt+

r∫

r0

crk−1+ dr for r > r0

≤ constant + crk+δ/(k+ δ) for r > r0.

This implies that order ofN(r) is not greater than that ofn(r), and ifn(r) has got mean or minimal type so hasN(R). The result for conver-gence or divergence class follows from earlier inequalities. The estimate 40

for n(r) in terms ofN(r) follows from

n(r) ≤ (1/ log 2)

2r∫

r

n(t)dtt≤ N(2r)/ log 2.

From this inequality it can be derived in the same way as before, thatthe order ofn(r) is not greater than that ofN(r) etc.

Definition . The order and type of n(r) or N(r) (being the same) arecalled the order and type of the sequence(an). The order of n(r) is alsosometimes calledexponent of convergenceof the sequence.

We note that∑

1/rkn converges if and only ifn(r) has order less than

k, or orderk of convergence type. (This is a consequence of lemma 3)

Next let us state the theorem,

Theorem 6. Suppose that an is a sequence having at most order q+ 1

(a positive integer) convergence class. Then the productπ(z) =∞∏

n=1E(z/an, q) converges absolutely and uniformly in any bounded region,and for |z| = r,

log |π(z)| < A(q)

rq

r∫

0

n(t)dt

tq+1+ rq+1

∞∫

0

n(t)dt

tq+2

36 Basic Theory

Proof.

E(u, q) = (1− u)eu+ 12u2+···+ uq

q

log E(u, q) = log(1− u) + u+12

u2+ · · · +

uq

q

= −∑

q+1

uk/k

so that| logE(u, q)| −∑

q+1

|u|k

k≤ 2|u|q+1 if |u| ≤

12

since log|E(u, q)| is the41

real part of logE(u, q) we get log|E(u, q)| ≤ q + 1 if |u| ≤12

. Suppose

12≤ |u| ≤ 1. Then

log |E(u, q)| ≤ log |(1− u)| + |u| +12|u|2 + · · · +

|u|q

q

≤ |u| + |u| +12|u|2 + · · · +

|u|q

q.

Also |u| ≥12

, |u|q−1 ≥ 1/2q−1, 2q−1|u|q−1 ≥ |u|. So that since|u| ≤ 1,

|u|q ≤ |u| ≤ 2q−1|u|q and

log |E(u, q)| ≤ 2q−1|u|q + q2q−1|u|q = A(q)|u|q < 2A(q).

A(q) being a constant depending only onq.Thus for|u| ≤ 1, log|E(u, q)| ≤ A(q)|u|q+1. �

Let

|u| ≥ 1, then log|E(u, q)| ≤ |u| + |u| +12|u|2 + · · · +

uq

q

≤ (q+ 1)|u|q

Now u = z/zn. Let |z| = r, |zn| = rn andN the least integer for whichrn ≥ r. Then|u| ≥ 1. Thus

N−1∑

n=1

log |E(z/zn, q)| ≤ A(q)rqN−1∑

1

r−qn

Basic Theory 37

and∞∑

n+1

logE(z/zn, q) ≤ B(q)rq+1∞∑

n+1

r−q−1n

Thus

∞∑

1

log |E(z/zn, q)| ≤ C

rqN−1∑

1

r−qn + rq+1

∞∑

N

r−q−1n

= C

rq

r∫

0

1tq

dn(t) + rq+1

∞∫

r

1

tq+1dn(t)

= K

rq

r∫

0

1tq+1

n(t)dt + rq+1

∞∫

r

1tq+2

n(t)dt

K depending only onq. This proves the theorem. 42

We also see that for largen,

∣

∣

∣

∣

∣

∣

logE

(

zzn, q

)∣

∣

∣

∣

∣

∣

< A(q)

(

rrn

)q+1

,

and, the product converges since∑

r−(q+1)n converges. Our Theorem 6

shows that if in Theorem 5f (z) has at most orderq convergence class,then the two products converge separately, uniformly and absolutely onevery bounded set.

As a consequence of the theorem we have

Theorem 7. If a sequence an defined as in the last theorem, has order

ρ, q− 1 ≤ ρ < q, q an integer, thenπa(z) =n=∞∏

n=1E(z/an, q− 1) has order

ρ and further ifρ is not an integer,∏

a(z) has the same type class as(an).

Hence if f (z) is meromorphic of finite non-integral order, then theroots of f (z) = a, have the same order and type class asf (z) except forat most one value ofa on the Riemann sphere.

38 Basic Theory

Proof. If n(t) < ctρ+ǫ for t > t0 whereǫ > 0

rq−1

r∫

0

n(t)dttq≤ rq−1

t0∫

0

n(t)dt/tq + rq−1

r∫

t0

ctρ+ǫdt/tq

− 0(rq−1) +Crρ+1−q+ǫ

ρ + 1+ −q+ ǫrq−1= 0

(

rq−1+

Crρ+ǫ

ρ + ǫ + 1− q

)

and the result regarding the order of the first integral follows. The43

second integral is treated similarly and then the first part follows fromTheorem 6.

If ρ > q− 1 and f (z) is of mean type or minimal type of orderρ wecan takeǫ = 0, and in the case of minimal typec small and the resultsfor the type at once follow. Suppose for instance thatn(t) has order lessthanq so thatn(t) < crρ+ǫ for t > t0. Then for larger,

r

∞∫

r

n(t)dt

tq+1< crq

∞∫

r

tρ+ǫ

tq+1dt =

crρ+ǫ

(q− ρ − ǫ)if ǫ < q− ρ.

and the integral is 0(rρ+ǫ ) and is 0(rρ) if n(r) = 0(rρ) as required. �

Suppose now thatf (z) is a meromorphic function of finite non-integral orderρ andq such thatq− 1 < ρ < q. Then,

f (z) = ePq−1(z)Π1(z)/Π2(z)

whereΠ1(z) andΠ2(z) are the products over the zeros and poles respec-tively. Now if both have a smaller order and type thanf (z), so does theirratio. since44

T(r,Π1/Π2) ≤ T(r,Π1) + T(r, 1/Π2)

= T(r,Π1) + T(r,Π2) + 0(1).

andePq−1(z) has orderq− 1 < ρ, so we should get a contradiction. Henceat least one of theΠ1(z) or Π2(z) and so either zeros or the poles havethe same order and type as the functionf (z).

Basic Theory 39

If for instance the poles do not have the order and type off (z) thensince f (z) − a has the same poles asf (z) (for every finitea), the roots off (z) = a that is the zeros off (z) − a have the order and type off (z).

This kind of argument was used to show that Riemann-Zeta functionhas an infinity of zeros.

We now illustrate the above by means of an example.If f (z) is meromorphic in the plane and lim

µ→∞

T(r, f )/ log r<+∞ then

f (z) is rational.In this casef (z) has lower order zero, and lower limit ofN(r, f )/

log r asr tends to∞ is less than∞. [T(r, f ) ≥ N(r, f )].Similarly,

limν→∞

N(r, 1/ f )log r

< ∞

further,

N(R, f ) ≥

R∫

r

n(t, f )dtt≥ n(r, f ) log

Rr

so thatn(r, f ) ≤N(r2, f )12 log r

= 0(1) for a sequence ofn → ∞. So f (z)

has only a finite number of zeros and poles. Hence from theorem5 45

f (z) = zp(Π(1 − z/aµ)/(1 − z/bν) where both the products are finite.Thus for all transcendental meromorphic functionsT(r, f )/ log r tendsto infinity asr tends to infinity.

Part II

Nevanlinna’s SecondFundamental Theorem

2.146

As the name indicates this theorem is most fundamental in thestudyof meromorphic functions. It is an extension of Picard’s Theorem, butgoes much farther. We develop the theorem in theorems 8 and 9 of thischapter, and then proceed systematically to explore some ofits conse-quences.

Theorem 8. Suppose that f(z) is a non-constant meromorphic functionin |z| ≤ r. Let a1, a2, . . . , aq be distinct finite complex numbers,δ > 0such that|aµ − aν| ≥ δ for 1 ≤ µ ≤ ν ≤ q. Then,

m(r, f ) +ν=q∑

ν=1

m(r, aν) ≤ 2T(r, f ) − N1(r) + S(r)

where N1(r) is positive and is given by

N1(r) = N(r, 1/ f ′) + 2N(r, f ) − N(r, f ′)

and

S(r) = m(r, f ′/ f ) +m

r,νq∑

ν=1

f ′/( f − aν)

41

Nevanlinna’s Second Fundamental Theorem 43

since all the term forµ , ν are at most log+ 2/δ. This is true if| f (z) −aν| < δ/3q for someν ≤ q. This inequality is true evidently for at mostoneν (with the condition|aµ − aν| ≥ δ). If it is not true for any valuethen we have trivially,

log+ |F(z)| ≥q

∑

ν=1

log+ 1/ f (z) − aν| − q log+ 3q/δ − log 2

(because, log+ |F(z)| ≥ 0). So the last relationship holds good in allcases. �

Taking integrals we deduce,

(2.1) m(r, F) ≥q

∑

ν=1

m(r, 1/ f − aν) − q log+3qδ− log 2

Again to get an inequality in the other direction,

m(r, F) = m

(

r,1f

ff ′

f ′F

)

By equation (1.2′) of 1.2 i.e. ,

T(r, f ) = T(r, 1/ f ) + log | f (0)|

m(r, f / f ′) = m(r, f ′/ f ) + N(r, f ′/ f ) − N(r, f / f ′) + log[| f (0)|/| f ′(0)|]

m(r, 1/ f ) = T(r, f ) − N(r, 1/ f ) + log 1/| f (0)|

Hence we get finally, 48

m(r, F) ≤ T(r, f ) − N(r, 1/ f ) + log 1/| f (0)| +m(r, f ′/ f )

+ N(r, f ′/ f ) − N(r, f / f ′) + log | f (0)|/| f ′(0)| +m(r, f ′F)

The above inequality, combined with (2.1) givesq∑

ν=1m(r, aν) − q log+

3q/δ − log 2≤ Right hand side of the above inequality. Add to both thesidesm(r, f ), we get the inequality,

m(r, f ) +q

∑

ν=1

m(r, aν) ≤ T(r, f ) + [m(r, f ) + N(r, f )] − N(r, f )


+m(r, f ′/ f ) + log 1/| f ′(0)| + log 2

+m

r,q

∑

ν=1

f ′/( f − aν)

+ q log+ 3q/δ − N(r, 1/ f )

+ N(r, f ′/ f ) − N(r, f / f ′).

= 2T(r, f ) − N1(r) +m(r, f ′/ f ) + log2| f ′(0)|

+ q log+3qδ+m

r,q

∑

ν=1

f ′

( f , aν)

where,N1(r) = N(r, f ) + N(r, 1/ f ) + N(r, f / f ′) − N(r, f ′/ f ). Now forany two functionsf (z) andg(z) Jensen’s formula gives

N(r, f /g) − N(r, g/ f ) =12π

2π∫

0

log

∣

∣

∣

∣

∣

∣

g(reiθ)f (reiθ)

∣

∣

∣

∣

∣

∣

dθ − log |g(0)/ f (0)|

=12π

2π∫

0

log |g(reiθ)|dθ − log |g(0)|

+12π

2π∫

0

log1

| f (reiθ)|dθ + log | f (0)|

= N(r, 1/g) − N(r, g) + N(r, f ) − N(r, 1/ f )

Thus49

N1(r) = N(r, f ) + N(r, 1/ f ) + N(r, 1/ f ′)

+ N(r, f ) − N(r, f ′) − N(r, 1/ f )

= 2N(r, f ) + N(r, 1/ f ′) − N(r, f ′)

as required.

2.2 Estimation of the error term

We shall firstly prove some lemmas.


Lemma 1. Suppose that f(z) is meromorphic in the|z| ≤ R and that0 < r < R. Letρ = 1

2(r + R) andδ(z) the distance of z from the nearestpole or zero of f(z) in |z| < ρ. Then,

m(r, f ′/ f ) < log+ T(R, f ) + 2 log+[1/(R− r)] + 2 log+ R

+12π

2π∫

0

log+1

δ(reiθ)dθ + 0(1)

0(1) depending only on the behaviour of f(z) at z= 0.

Proof. We have by differentiation of Poisson formula as in theorem 5,

f ′(z)/ f (z) =12π

2π∫

0

log | f (ρeiφ)|2ρeiφ

(ρeiφ − z)2dφ

+

∑

µ

[

aµ(ρ2 − aµz)

−1

(aµ − z)

]

+

∑

ν

1(bν − z)

−bν

(ρ2 − bνz)

where the sums as usual run over the zerosaµ and polesbν in |z| < ρ.From |ρ2 − aµz| ≥ ρ2 − ρ|z| = ρ2 − rρ = (ρ − r)ρ for |z| = r we get|aµ|

|ρ2 − aµz|≤

ρ

ρ2 − ρr=

1(ρ − r)

and by definition ofδ(z)

∣

∣

∣

∣

∣

∣

1(aµ − z)

∣

∣

∣

∣

∣

∣

≤1δ(z)

,

∣

∣

∣

∣

∣

1bν − z

∣

∣

∣

∣

∣

≤1δ(z)

Hence 50∣

∣

∣

∣

∣

∣

∑

[

aµ(ρ2 − aµz)

−1

(aµ − z)

]

+

∑

1(bν − z)

−bν

ρ2 − bνz

∣

∣

∣

∣

∣

∣

≤ [n(ρ, f ) + n(ρ, 1/ f )]

[

1δ(z)+

1(ρ − r)

]

We now estimaten(ρ, f ) andn(ρ, 1/ f ). For this we have,

R∫

ρ

n(t, f )dt/t ≤ N(R, f ) ≤ T(R, f )


and sincen(t, f )/t ≥ n(ρ, f )/R for t greater thanρ,R∫

ρ

n(t, f )dt/t ≥

n(ρ, f )(R− f )/R. Therefore [n(ρ, f )](R− ρ)/R ≤ T(R, f ) or n(ρ, f ) ≤RT(R, f )/(R− ρ) = 2RT(R, f )/(R− r) since 2ρ = (R+ r).

Similarly n(ρ, 1/ f ) ≤ 2RT(R, 1/ f )/(R−r) = 2R[T(R, f )+0(1)]/(R−r). 0(1) depends only on the behaviour off (z) at z= 0. Thus

n(ρ, f ) + n(ρ, 1/ f ) ≤4R

(R− r)[T(R, f ) + 0(1)]

and so∣

∣

∣

∣

∣

∣

∑

[

aµρ2 − aµz

−1

(aµ − z)

]

+

∑

1(bν − z)

−bν

(ρ2 − bνz)

∣

∣

∣

∣

∣

∣

≤4R

(R− r)[T(R, f ) + 0(1)]

[

1δ(z)+

2(R− r)

]

Further,

12π

∣

∣

∣

∣

∣

∣

∣

∣

∣

2π∫

0

log | f (ρeiφ)|2ρeiφ

(ρeiφ − z)2dφ

∣

∣

∣

∣

∣

∣

∣

∣

∣

≤12π

2ρ

(ρ − r)2

2π∫

0

log | f (ρeiφ)|dφ

for,51

|ρeiφ − z| ≥ ||ρeiφ | − |z|| = (ρ − r)

=1π

4ρ

(R− r)2

2π∫

0

log+ | f (ρeiφ)dφ +

2π∫

0

log+1

| f (eiφρ)|dφ

=8ρ

(R− r)2[m(ρ, f ) +m(ρ, 1/ f )] ≤

8ρ

(R− r)2[2T(ρ, f ) + 0(1)]

=16ρ

(R− r)2[T(ρ, f ) + 0(1)] ≤

16R

(R− r)2[T(R, f ) + 0(1)]


0(1) depending only on the behaviour off (2) atz = 0. Thus from theabove inequalities and the equation forf ′(z)/ f (z) we get finally,

| f ′(z)/ f (z)| ≤4R

(R− r)[T(R, f ) + 0(1)]

[

2(R− r)

+1δ(z)

]

+16R

(R− r)2[T(R, f ) + 0(1)]

=4R

(R− r)2

[

6T(R, f ) + T(R, f )(R− r)δ(z)

+ 0(1)+ 0(1)R− rδ(z)

]

≤4R

(R− r)2

[

6+R− rδ(z)

]

[T(R, f ) + 0(1)]

Hence,

log+ | f ′(z)/ f (z)| ≤ log+4R

(R− r)2+ log+

(

6+R− rδ(z)

)

+ log+ T(R, f )

+ 0(1)≤ log+ R2 log+1

(R− r)+ log+ T(R, f ) log+

(R− r)δ(z)

+ 0(1)

≤ 2 log+ R+ log+1δ(z)+ 2 log+

1(R− r)

+ log+ T(R, f ) + 0(1)

0(1) depending only on the behaviour off (z) at z = 0. Integrating the 52

above inequality on the circle|z| = r,

12π

2π∫

0

log | f ′(reiθ)/ f (reiθ)|dθ ≤ 2 log+ R+12π

2π∫

0

log+1

δ(reiθ)dθ

+2 log+1

(R− r)+ log+ T(R, f ) + 0(1)

which gives the lemma. �

Lemma 2. Let z be any complex number and0 < r < ∞. Let Ek be theset of allθ such that|z− reiθ | < kr where0 < k < 1. Then

∫

Ek

log[r/|(z− reiθ)|]dθ < π

[

log

(

1k

)

+ 1

]


Proof. We may after rotation assumez real and positive. For ifz= 0 Ek

is obviously void and there is nothing to prove. So letz> 0. Then forθin Ek |z− reiθ | ≥ | Im(z− reiθ)| = r sinθ andEk is contained in an intervalof the form|θ| < θ0 wherer sinθ0 ≤ kr, that is, sinθ0 ≤ k. So,

∫

Ek

log[r/|reiθ − z|]dθ ≤ 2

θ0∫

θ

log+1

Sinθdθ

Now, θ <π

2for when

π

2≤ |θ| ≤ π, |z− reiθ | > |reiθ | = r. Thus53

sinθθ≥

2π|θ| < θ0 we get

∫

Ek

log[r/|reiθ − z|]dθ ≤ 2

θ0∫

0

logπ

2θdθ

= 2

θ0∫

0

logπ

2θdθ

becauseθ0 <π

2or

π

2θ> 1.

∫

Ek

log[r/|(reiθ − z)|]dθ ≤ 2

θ0∫

0

logπ

2dθ − 2

θ0∫

0

logθdθ

= 2θ0 logπ

2− 2θ0 logθ0 + 2θ0

�

Lemma 3. With the hypothesis of lemma 1,

m(r, f ′/ f ) ≤ 3 log+ T(R, f )+4 log+R+4 log+[1/(R−r)]+log+(1/r)+0(1)

Proof. Notations being the same as in the proof of lemma 1, write,

[1/δ(z)] = [r/δ(z)] · (1/r). Then log1δ(z)≤ log

1r− log

rδ(z)

. So

12π

2π∫

0

log+1

δ(reiθ)dθ ≤ log+

1r+

12π

2π∫

0

log+r

δ(reiθ)dθ


Let E denote the set ofθ in (0, 2π) for which δ(reiθ) <r

n2wheren =

n(ρ, f ) + n(ρ, 1/ f ). (If n = 0 so that there are no zeros and poles we canputδ(z) = +∞ and there is nothing to prove. So assumen ≥ 1). For eachpointθ in E there is a zero or polezν such thatδ(reiθ) = |zν− reiθ | < r/n2 54

and thenlog+[r/δ(z)] = log+[r/|reiθ − zν|]

Now write log0 x = log x if x ≥ n2 and log0 x = 0 otherwise. Then sincen ≥ 1, log0 x ≥ 0 always.

Also for θ in E log+[r/δ(z)] = log+[r/|(reiθ − zν)|] for someν Since,1 < n2 < [r/|reiθ − zν|],

log+r

|reiθ − zν|= log0

r

|reiθ − zν|≤

∑

µ

log0r

|reiθ − z|

where the sum is taken over all zeros and poleszµ in |z| < ρ. Thus∫

E

log+r

δ(reiθ)dθ ≤

∑

µ

∫

E

log0r

|reiθ − zµ|dθ

≤∑

µ

π

n2[log n2

+ 1] =π

n[2 logn+ 1] ≤ A.

by lemma 2 withk = 1/n2, and whereA is an absolute constant.Also on the complement ofE, δ(reiθ) ≥ r/n2 and,

∫

compl. ofE

log+r

δ(reiθ)dθ ≤

2π∫

0

logn2dθ = 2π logn2

Adding we obtain

12π

2π∫

0

log+r

δ(reiθ)dθ ≤ [2 logn+ A]

12π

2π∫

0

log+1

|δ(reiθ)|dθ =

12π

2π∫

0

log+1r

r

|δ(reiθ)|dθ


≤12π

2π∫

0

logr

|δ(reiθ)|dθ + log+

1r

≤ 2 log+ n+ log+ 1/r + A.

wheren = n(ρ, f ) + n(ρ, 1/ f ).55

We have from (2.2)

n = n(ρ, f ) + n(ρ, 1/ f ) ≤ 4R[T(R, f ) + 0(1)]/(R− r).

Hence,

log+ n ≤ log+ R+ log+ 1/(R− r) + log+ T(R, f ) + 0(1),

giving

12π

2π∫

0

log1

|δ(reiθ)|dθ ≤ log+

1r+ 2 log+

1R− r

+ 2 log+ R

+ 2 log+ T(R, f ) + 0(1)

From this and lemma 1, lemma (3) follows. That is,

m(r, f ′/ f ) < 3 log+ T(R, f )+ 4 log+ 1/(R− r)+ 4 log+R+ log+ 1/r 0(1).

�

Lemma 4 (Borel). (i) Suppose T(r) is continuous, increasing andT(r) ≥ 1 for r0 ≤ r < ∞. Then we have

(2.3) T[r + 1/T(r)] < 2T(r)

outside a set E of r which has length (that is) linear measure atmost2.

(ii) If T (r) is continuous and increasing for r0 ≤ r < 1 and T(r) ≥ 156

then we have

(2.4) T[r + (1− r)/(eT(r))] < 2T(r)


outside a set E of r such that∫

E

dr/(1 − r) ≤ 2. In particular

T[r + (1 − r)/eT(r)] < 2T(r) for some r such thatρ < r < ρ′ ifr0 < ρ < 1 and1− ρ′ < (1− ρ)/e2.

Proof. We prove first (i) [that is in the plane]. Letr1 be the lower boundof all r for which (2.3) is false. If there are no suchr there is nothing toprove. We now define by induction a sequence of numbersrn. Supposethatrn has been defined and writer′n = rn+ 1/T(rn). Define thenrn+1 asthe lower bound of allr ≥ r′n for which (2.3) is false. We have alreadydefinedr1 and so we obtain the sequence (rn). Note that by continuity ofT(r) (2.3) is false forr = rn, for n = 1, 2, 3, . . . that isrn belongs toE0,E0 being the exceptional set. From the definition ofrn+1 is follows thatthere are no points ofE in (r′n, rn+1) and that the set of closed intervals[rn, r′n] containsE0. If there are an infinity ofrn, (rn) cannot tend to afinite limit r. For then sincern < r′n ≤ rn+1, r′n tends tor also. Butr′n − rn = 1/T(rn) which is greater or equal to 1/T(r) > 0, sinceT(r) isincreasing, for allmwhich is a contradiction. �

It remains to be shown that∑

(r′n − rn) ≤ 2. Now T(r′n) = T[rn +

1/T(rn)] ≥ 2T(rn) sincern belongs toE. And soT(rn+1) ≥ T(r′n) ≥2T(rn). Therefore,T(rn+1) ≥ 2T(rn) ≥ . . . ≥ 2nT(r1) ≥ 2n sinceT(r) ≥ 57

1. ThusT(rn) ≥ 2n−1. Now∑

(r′n − rn) =∑

1/T(rn) ≤∑

21−n ≤ 2.To prove part (ii) of the theorem, setR log[1/(1− r)] gettingr = 1−

e−R and putT(r) = ϕ(R)·ϕ(R) then is continuous and increasing forR0 =

log[1/(1− r0)] ≤ R< ∞ andϕ(R) ≥ 1. Apply then the first part toϕ(R).Then we haveϕ[R+1/ϕ(R)] < 2ϕ(R) for R> log[1/(1−r0)] outside a setE of Rsuch that 2≥

∑

(R′n−Rn) =∫

E

dR=∫

dr/(1−r). Translating back

to r, R′ = R+ 1/ϕ(R) becomes log[1/(1− r′)] = log[1/(1− r)] + 1/T(r).

That is (1− r′) = (1 − r)e−1

T(r) andT(r′) < 2T(r). By the first meanvalue theoremf (b) = f (a) + (b− a) f ′(x), wherea ≤ x ≤ b. SinceT(r)increases withr, T[r + (1 − r)/eT(r)] < 2T(r) outside the exceptionalsetE of r for which

∫

E

dr/(1 − r) ≤ 2. If E contains the whole of the

interval ρ < r < ρ′ then∫ ρ′

ρdr/(1 − r) ≤

∫

E

dr/(1 − r) ≤ 2, that is

log(1− ρ)/(1− ρ′) ≤ 2, and so (1− ρ)/(1− ρ′) ≤ e2 as required.


2.3

Theorem 9. If f (z) is meromorphic and non-constant in the plane andS(r, f ) denotes the error term in Theorem 8 then we have

(2.5) S(r, f ) = O[log T(r, f )] +O(log r)

as r tends to infinity through all values if f(z) has finite order, andthrough all values outside a set of finite linear measure otherwise.

(ii) If f (z) is meromorphic (non-constant) in|z| < 1 and thelimr→1{T(r, f )/ log[1/(1 − r)]} = ∞, then we have S(r, f ) = O[T(r, f )]

as r tends to one on a set E such that∫

E

dr/(1− r) = ∞.

Proof. If ϕ(z) =q∏

ν=1[ f (z) − aν] then

S(r, f ) = m(r, f ′/ f ) +m[r, ϕ′/ϕ] +O(1)

becauseS(r, f ) = m(r, f ′/ f )+m(r,q∑

ν=1f ′/( f −aν))+O(1) and

q∑

ν=1f ′/( f −58

aν) =ϕ′

ϕby logarithmic differentiation. Therefore from the lemma 3,

for anyR> r we have,

S(r, f ) ≤ 3 log+ T(R, f ) + 4 log+R+ 4 log+[1/(R− r)]

+ log+(1/r) + 3 log+ T(R, ϕ)

4 log+ R+ 4 log+ 1/(R− r) + log+(1/r) + 0(1)

Also

(2.6) T(r, ϕ) ≤q

∑

ν=1

T(r, f − aν) ≤ T(r, f ) +O(1), . . .

Thus

S(r, f ) ≤ 3(1+ q) log+ T(R, f ) + 8 log+ R+ 8 log+[1/(R− r)]

+2 log+(1/r) +O(1)


If r is greater than 1 (we are consideringS(r, f ) for larger) log+(1/r) =0. Now suppose thatf (z) is meromorphic of finite order so thatT(r, f ) <rK for r greater thanr0. Also chooseR= r2 andr ≥ 2 so thatR− r > 1.

Then log+(

1R− r

)

= 0. We get

S(r, f ) ≤ 3(1+ q) log+ T(R, f ) + 8 log+R+O(1),

log+ T(R, f ) < k log+R< 2k log+ r = 2k log r

sinceR= r2 andr > 1. We thus have finally,S(r, f ) ≤ 6(1+ q)K log r +16 logr + O(1) showing thatS(r, f ) = 0(logr) which gives (2.5) sincelog+ T(r, f ) = O(log r). �

Note that by our examplesT(r, f )/ log r tends to infinity unlessf (z) 59

is a rational function in which caseS(r, f ) is bounded becausef ′/ f → 0asz→ ∞ for any polynomial and hence for any rational function. Thusif f (z) has finite orderS(r, f )/T(r, f ) → 0 asr → ∞.

If f (z) has infinite order takeR = [r + 1/T(r)], then log+ R ∼log r[sinceT(r) → ∞] · log+ 1/(R − r) = log+ T(r) and log 1/r = 0finally. Outside the exceptional set of lemma 4,T(R, f ) < 2T(r, f ) andso, log+ T(R, f ) ≤ log+ T(r, f ) + log 2. Hence again we have (2.5) out-side the exceptional set. This completely proves i) In orderto prove

(ii) denote byrn a sequence such thatT(rn, f )/ log

(

11− rn

)

→ ∞ as

n → ∞ and by taking a sub sequence if necessary we can assume that

1− rn+1 <1− rn

10. Then letrn be defined by 1− r′n = (1− rn)/10 so that

rn < r′n < rn+1 < 1. Then sincer ′n∫

rn

dr1− r

= log(1− rn)/(1− rn) = log 10>

2 the unionE1 of all the intervals (rn, r′n) is such that∫

E1

dr1− r

= +∞.

Further each such interval contains a point not in the excep-tional set E, for T(r, f ) because by lemma 4

∫

E

dr/(1 − r) ≤ 2, pro-

vided only thatT(r1) > 1. For a not exceptional pointr of E1 take


R= r + (1− r)/eT(r), then

log+1

R− r= log+

eT(r)1− r

< log+ T(r) + log+1

1− r+ 1

and log+ T(R) < log+ T(r) + log 2, Thus by (2.6)60

S(r, f ) < 3(1+ q) log+ T(R, f ) + 8 log+R+ 8 log+1

(R− r)+O(1)

< (3+ 3q+ 8) log+ T(r) + 8 log+1

(1− r)+O(1)

Also, log+[1/(1− r)] < log+ 1/(1− r′n)+ log+ 10(1−rn) < log+ 1

1−rn+0(1)=

O[T(rn)] +O(1) = O[T(r)].So since logT(r, f ) = O T(r) we getS(r, f ) = O T(r). This proves

(ii) for a setE1 of r such that∫

E1

dr1− r

= +∞ and containing at least one

point in each intervalrn < r < r′n. In faceE comprises all ther in thesequence of intervals [rn, r′n], n > 1 except possibly a setE0 such that∫

E0

dr1− r

≤ 2.

2.4 Applications61

Definition . Let n(t, a) denote the number of roots of f(z) = a in |z| ≤t, the multiple roots being counted according to their multiplicity andn(t, a) the number of roots of f(z) = a in |z| < t with the multiple roots

counted simply. Further letN(t, a) =r∫

0

n(t, a) − n(0, a)dt/t. [ n(0, a) is

equal to one if f(0) = a and zero otherwise]; and N(r, a) as before withn(t, a) instead ofn(t, a). Let now the function f(z) be meromorphic, andnon-constant in|z| < R, 0 < R ≤ ∞ and suppose that f(z) satisfies thehypotheses of theorem 9, so thatlim

r→R[S(r, f )/T(r)] = 0 and T(r) tends

to infinity as r tends to R.

Now writeδ(a) = limr→R

[m(r, a)/T(r)] = 1− limr→R

[N(r, a)/T(r)] because

[m(r, a) + N(r, a)]/T(r) = [T(r) + 0(1)]/T(r) → 1;


thus

limr→R

[m(r, a)/T(r)] = limr→R

1+−N(r, a)

T(r)= 1+ lim

r→R−N(r, a)/T(r)

= 1− limr→R

N(r, a)/T(r)

Again write,θ(a) = lim

r→R[N(r, a) − N(r, a)]/T(r)

andΘ(a) = 1− lim

r→RN(r, a)/T(r) = lim

r→R[1 − N(r, a)/T(r)].

Clearly,δ(a), θ(a) andΘ(a) lie in the closed interval [0, 1]. Also

1−NT= 1−

NT+

N − NT

limr→R

1−NT≥ lim

r→R

(

1−NT

)

+ limr→R

N − NT

i.e.Θ(a) ≥ δ(a) + θ(a). 62

The quantityδ(a) is called thedefect of aand θ(a) the Branchingindex (Verzweigungsindex) ofa. Now we have the defect relation ofNevanlinna. This is the second fundamental theorem in its most effec-tive form and is very important in the theory.

Theorem 10. If f (z) satisfies the hypotheses of the Theorem 9, thenΘ(a) = 0 except possibly for a finite or countable sequence aν of valuesof a and for these

∑

Θ(aν) ≤ 2.

Proof. From theorem 8, for anyq distinct valuesaν, of a includinga1 =

∞q

∑

ν=1

m(r, aν) < 2T(r, f ) − N1(r) + S(r)

and addingq∑

ν=1N(r, aν) to both sides and using first fundamental theorem

T(r, a) = T(r, f ) +O(1), we get

qT(r, f ) < 2T(r, f ) − N1(r) +q

∑

ν=1

N(r, aν) + S(r, f ) + 0(1)


(q− 2)T(r, f ) <q

∑

ν=1

N(r, aν) − N1(r) + S(r) + 0(1).

Now, N1(r) = 2N(r, f ) − N(r, f ′) + N(r, 1/ f ′), and since by definitionN(r, f ) = N(r,∞), N(r,∞) − N1(r) = N(r, a1) − N1(r) = N(r, f ′) −N(r, f ) − N(r, 1/ f ′).

So,63

(q−2)T(r, f ) <q

∑

ν=1

N(r, aν)+N(r, f ′)−N(r, f )−N(r, 1/ f ′)+S(r)+0(1)

If f (z) has a pole of orderp at z0, f ′(z) has a pole of orderp + 1 atz0

so thatn(t, f ′) − n(t, f ) = n(t, f ) and soN(r, f ′) − N(r, f ) = N(r, f ),.Similarly if a is anyone ofa2, a3, . . . , aq (finite) and f (z) = a has a rootof multiplicity p, f ′(z) has there a zero of orderp− 1. Thus

q∑

ν=2

N(r, aν) − N(r, 1/ f ′) =q

∑

ν=2

N(r, aν) − N0(r, 1/ f ′)

whereN0(r, 1/ f ′) refers to zeros off ′(z) at points other than the rootsof f (z) = a.

Hence we get,

(q− 2)T(r, f ) <q

∑

ν=2

N(r, aν) − N0(r, 1/ f ′) + S(r) + N(r,∞) + 0(1)

i.e. (q− 2)T(r, f ) <q

∑

ν=1

N(r, aν) − N0(r, 1/ f ′) + S(r, f ) + 0(1)

i.e.q

∑

ν=1

N(r, aν)T(r)

≥ (q− 2)−S(r, f ) + 0(1)

T(r)since N0(r, 1/ f ′) ≥ 0

since limr→R

S(r, f )/T(r) = 0 andT(r, f )→ ∞ asr → R

limr→R

[

−O(1)+ S(r)

T(r)

]

= 0. Thus limr→R

q∑

ν=1

N(r, aν)T(r)

≥ (q− 2)


and afortioriq

∑

ν=1

limr→R

N(r, aν)T(r)

≥ q− 2

that is, 64q

∑

ν=1

[1 − Θ(aν)] ≥ (q− 2) as required.

This shows thatΘ(a) > (1/n) at most for 2n values ofa and soΘ(a) > 0for at most a countable number ofa. If thesea’s are arranged in a

sequenceq∑

r=1Θ(ar) ≤ 2 for any finiteq, and so to infinity. �

Consequences

(1) Sinceδ(a) ≤ Θ(a) we have∑

δ(aν) ≤ 2 and thus there existsat most two values ofa for which δ(a) = 1, or more generallyδ(a) > 2

3. Thus if the equationf (z) = a has only a finite numberof roots in the plane,N(r, a) = 0(logr) asr tends to infinity, andwe should have

limr→∞

log[r/T(r)] > 0, i.e. limr→∞

T(r)log r

< ∞.

i.e., f (z) is rational. Thus if f (z) is transcendental in the planeδ(a) < 1 the equationf (z) = a has infinitely many roots. Thesame is true in all cases ifR is finite, since otherwiseN(r, a) =0(1) and so lim

r→RT(r) < +∞ that isT(r) = 0(1) asr tends toR,

sinceT(r) is increasing. This would contradict

limr→R

T(r)/ log

(

11− r

)

= +∞.

This result thus contains Picard’s theorem as a special case.

(2) Θ in relation to N andN. Suppose that the functionf (z) = a has 65

only multiple roots of multiplicityk ≥ 2. Then

N(r, a) ≤ (1/k)N(r, a) ≤ (1/k)[T(r, f ) + 0(1)]


Hence in this case,

limr→R

N(r, a)/T(r) ≤ (1/k)

Θ(a) = 1− limr→R

N(r, a)/T(r) ≥ 1− (1/k) ≥12

If a1, a2, . . . , aq areq such values withk = kν for aν, we havesince

∑

νΘ(aν) ≤ 2,

∑

ν[1 − (1/kν)] ≤ 2.

In particular there can be at the most four such valuesaν of a for[1 − (1/kν)] ≥ 1

2.If f (z) is regularm(r, f ) = T(r, f ) andδ(∞) = limm(r, f )/T(r, f ) =

1. and soΘ(∞) = 1 becauseΘ(∞) ≤ δ(∞) = 1. So,∑

νΘ(aν) ≤ 1 for

any finite number of finiteaν’s. So that there can be only two such val-uesaν which are taken multiply. Such values are called fully branched(Vollstandig Verzweight). These results are best possible, for sinz andcosz, have the values±1 “fully branched”. Again for the Wierstrassianelliptic functionP(z) which satisfies the differential equation

[P′(z)]2= (P(z) − e1)(P(z) − e2)(P(z) − e3)

wheree1, e2, e3 are distinct finite numbers the valuese1 e2 ande3 areevidently fully branched and so is infinity. IfP(z) has a pole of orderk,P(z) ∼ A(z−ζ)−k and [P′(z)]2 ∼ P(z)3 ∼ A3(z−ζ)−3k from the differential66

equation.But

[P′(z)]2 ∼ A′z−2k−2 i.e., − 2k − 2 = −3k

Hence we havek = 2 so all the poles are double and infinity is fullybranched.

We also note that the equationw2=

∏qν=1(z− aν) can have no para-

metric solutionz = ρ(t), w = ψ(t) which are integral functions oft ifq ≥ 3 or which are meromorphic ifq ≥ 5. Because ifϕ(t0) = a1 forinstance thenw2

= ψ2(t) has a zero att0 and soψ(t) has a zero andψ2(t)has at least a double zero att0. Hence alsoϕ(t)−a1 has at least a dou-ble zero att0 and all roots ofϕ(t) = aν therefore will be multiple and

q∑

ν=1Θ(aν) ≥ 2 for ϕ(t), a contradiction.


We just remark that this result extends to a general equationg(z,w) = 0 of genus greater than 1.

67

Theorem 11. Suppose f(z) is meromorphic of finite order in the planeandΘ(a1) = Θ(a2) = 1, a1 , a2. Then if a is not equal to a1, a2,N(r, a) ∼ T(r) as r tends to infinity.

Proof. In fact we have iff (z) is not rational

T(r, f ) < N(r, a1) + N(r, a2) + N(r, a) + S(r, f ) +O(1)

and in this caseS(r, f ) = O[T(r)], (evenS(r) = 0(logr)). By hypoth-esis sinceΘ(ai) = 1 − lim[N(r, a)/T(r)], N(r, ai ) = 0[T(r)] i = 1, 2.Therefore [1+ 0(1)]T(r, f ) < N(r, a) asr → ∞.

That is limr→∞

[N(r, a)/T(r)] ≥ 1, and evidentlylimr→∞

N(r, a)T(r)

≤ 1 that

is, N(r, a)/T(r) → 1, asr tends to infinity. Similarly sinceN(r, a) ≥N(r, a), lim

r→∞N(r, a)/T(r) ≥ 1 and againN(r, a) ∼ T(r). If f (z) is rational

these results follow by elementary methods; in fact in this case there isat most onea namelya = f (∞) for which N(r, a) = OT(r) unlessf is aconstant. �

Theorem 12. If aν(z) for ν = 1, 2, 3 are three functions satisfyingT(r, aν) = O[T(r, f )] as r → R, and f(z) is as in theorem 10, thenwe have

[1 +O(1)]T(r, f ) <3

∑

i=1

N

(

r,1

f − ai(z)

)

as r tends to R through a suitable sequence of values.

Proof. Set 68

ϕ(z) =f (z) − a1(z)f (z) − a3(z)

a2(z) − a3(z)a2(z) − a1(z)

It is easy to see usingT(r, ai ) = O[T(r, f )] that T(r, ) = T(r, f ) =[1+O(1)]T(r, f ), alsoϕ = 0, 1,∞, only if f − a1(z), f − a2(z), f − a3(z)


are zero or if two suitablea’s are equal.

N

[

r,1

a2 − a3

]

≤ T[r, 1/(a2 − a3)] = T[r, a2 − a3] +O(1)

≤ T(r, a2) + T(r, a3) +O(1) = O[T(r)]

So,N(r, 1/ϕ) ≤ N

(

r,1

f − a1

)

+O[T(r, f )] etc. by Jensen’s formula and

hypothesis. �

In the course of the proof of theorem 10 we obtained,

(q− 2)T(r, f ) <q

∑

i=1

N(r, ai ) − N0(r, 1/ f ′) + S(r, f ) +O(1)

<

q∑

i=1

N(r, ai ) + S(r, f ) +O(1).

Take f = ϕ, anda1 = ∞, a2 = 0, a3 = 1 to get

T(r, ϕ) < N(r, ϕ) + N

(

r,1ϕ

)

+ N

(

r,1

ϕ − 1

)

+O[T(r, ϕ)]

That is,

[1 +O(1)]T(r, ϕ) < N(r, ϕ) + N

(

r,1ϕ

)

+ N

(

r,1

ϕ − 1

)

for a suitable sequence ofr tending toR. ϕ = 0, only if either f = a1,

or a2 = a3. SoN

(

r,1ϕ

)

≤ N

(

r,1

f − a1

)

+ N

(

r,1

a2 − a3

)

which is equal69

to N

(

r,1

f − a1

)

+O[T(r)]. Similar reasoning givesN

(

r,1ϕ

)

+ N(r, ϕ) +

N

(

r,1

ϕ − 1

)

<3∑

i=1N

(

r,1

f − ai

)

+ O[T(r)]. Thus sinceT(r, ϕ) = [1 +

O(1)]T(r) the result follows.

Remark. Note that the same reasoning cannot be applied to more thanthree functions. In fact it is not known whether the analogous result isstill true if we take more than three functions.


2.5 Picard values of meromorphic functions and their deriva-tives:

70

A function f (z) will be called admissible if it satisfies the hypothesis oftheorem 9 in a circle|z| < Rand alsof (z) is transcendental ifR= ∞.

Note that if f (l)(z) lth derivative off (z), all the poles off (1)(z) havemultiplicity at leastl + 1. Therefore,N[r, f (l)] ≤ N(r, f (l))/(l + 1) and

Θ(∞) = lim

(

1−N(r, f (l))

T(r, f (l))≥ lim

)

1−1

(l + 1)N(r, f (l))

T(r, f (l))≥

l(l + 1)

We obtain for f (l)(z), sinceq∑

0Θ(aν) ≤ 2, if a1, . . . , aq, are distinct and

finite,q∑

1Θ(aν) ≤ 1 +

1l + 1

. Thus there can be at most one finite value

which is taken only a finite number of times or more generally for which

Θ(a) >34

.

Write nowψ(z) = a0(z) + f (z) + · · · + al(z) f l(z)ai (z) being functionssatisfyingT[r, ai (z)] = OT(r, f ) and we assumeψ(z) is not identicallyconstant. Then we have the following sharpened form of a theorem ofMilloux.

Theorem 13. If f (z) is admissible in|z| < R then,

T(r, f ) < N(r, f ) + N(r, 1/ f ) + N

[

r,1

ψ − 1

]

− N0(r, 1/ψ′) + S1(r, f )

where N0(r, 1/ψ′) indicates that zeros ofψ′ corresponding to the re-peated roots ofψ = 1 are to be omitted, andlim

r→RS1(r, f )/T(r, f ) = 0.

Note that this reduces to the second fundamental theorem forq = 3whenψ = f .

71

Firstly let us prove some lemmas.


Lemma 5. If l is a positive integer and f(z) admissible in|z| < R, such0 < r < ρ < R then,

m

[

r,f (l)

f

]

< A(l)

[

log+ T(ρ, f ) + log+ ρ + log+1

ρ − r+ log+(1/r) +O(1)

]

A(l) being a constant depending only on l.

Proof. The proof is by induction. The result is true forl = 1, by Lemma3 of section 2.1. Takeρ1 =

12(ρ + r) and assume that result forl. Now,

T[ρl , f 1(z)] = m[ρ1, f l(z)] + N[ρ1, f l(z)]

≤ m[ρ1, f ] +m(ρ1, f (l)/ f ) + (l + 1)N(ρ1, f )

since at a pole off of orderk, f (l) has a pole of orderk + l ≤ k(l + 1).By our induction hypothesis the right hand side is less than

(l + 1)T(ρ1, f ) + A(l)

[

log+ T(ρ, f ) + log+ ρ

+ log+1

ρ − ρ1+ log+

1ρ1+O(1)

]

< [l + 1+ A(l)]T(ρ, f ) + A(l)

[

log+ ρ + log+1

ρ − ρ1+ log+

1ρ1+O(1)

]

So,

log+ T[ρ1, f (l)] < log+ T(ρ, f ) + log+(log+ ρ) + log+(

log+1

ρ − ρ1

)

+ log+(

log+1ρ1

)

+O(1)

< log+ T(ρ, f ) + log+ ρ + log+1

ρ − ρ1+ log+

1ρ1+O(1)

Also by Lemma 3 applied tof (l),72

m

(

r,f l+1

f l

)

< 3 log+ T[

ρ1, f (l)]

+ 4 log+ ρ1 + log+1r

+ 4 log+1

ρ1 − r+O(1)


< 3 log+ T(ρ, f ) + 7 log+ ρ + 7 log+1

ρ − r+ 2 log+

1r+O(1)

< A

[

log+ T(ρ, f ) + log+ ρ + log+1

ρ − r+ log+

1r+O(1)

]

because,

ρ1 − r = ρ − ρ1 =12

(ρ − r)(ρ > ρ1 > r)

Therefore, since

m

[

r,f l+1

f

]

< m

[

r,f l+1

f (l)

]

+m[

r, f (l)/ f]

.

m

(

r,f l+1

f l

)

is less than

[A+ A(l)]

[

log+ T(ρ, f ) + log+1

ρ − r+ log+(1/r) +O(1)+ log+ ρ

]

completing the inductive proof. �

Lemma 6. If ψ(z) is defined as in theorem 13, and0 < r < ρ < R, then

m(r, ψ/ f ) < O[T(r, f )] +A(l)[log+ T(ρ, f )+ log+1

ρ − r+ log+ ρ] +O(1)

as r→ R in any manner.

Proof.

m(r, ψ/ f ) ≤l

∑

i=0

m

r, ai(z) f (i)

f

+ log(l + 1).

≤

l∑

i=0

m[r, ai(z)] +l

∑

i=0

m[r, f (i)/ f ] + log(l + 1)+ (l + 1) log 2.

= O[T(r)] +l

∑

i=0

m[r, f (i)/ f ] +O(1)

< O[T(r)] + A(l)

[


ρ − r+ log+ ρ

]

+O(1)

from the lemma 5 and becausem[r, ai(z)] ≤ T[r, ai (z)] = O[T(r)], and 73

log+(1/r) remains bounded asr tends toR. �


Lemma 7. If ψ(z) is defined as above then as r→ R in any mannerwhile 0 < r < ρ < R,

T(r, ψ) < [l + 1+O(1)]T(r, f ) + A(l)[

log+ T(ρ, f ) + log+(

1ρ − r

)

+ log+ ρ +O(1)

]

.

Proof. m(r, ψ) ≤ m(r, ψ/ f )+m(r, f ). If f has a pole of orderK at a pointandaν(z) a pole of orderKν thenaν(z) f (ν) has a pole of orderν+K+Kν,and soψ(z) has a pole or of order Max. (ν + K + Kν) ≤ (l + 1)K +

∑

Kν.This gives

N(r, ψ) ≤ (l + 1)N(r, f ) +∑

ν

N(r, aν)

≤ (l + 1)N(r, f ) +O[T(r, f )]

Adding the above two inequalities,

T(r, ψ) ≤ [l + 1+O(1)]T(r, f ) +m(r, ψ/ f ),

and now the lemma follows from the previous lemma. �

Lemma 8. If S(r, ψ) is defined as in Theorem 8 withψ instead of f thenif 0 < r < ρ < R and r tends to R,

S(r, ψ) < A

[

log+ T(ρ, f ) + log+(

1ρ − r

)

+ log+ ρ +O(1)

]

.

Proof. Let ρ1 =12(ρ + r). Lemma 3 gives then

S(r, ψ) < A

[

log+ T(ρ1, ψ) + log+(

1ρ1 − r

)

+ log+ ρ1 +O(1)

]

.

By lemma 7 since log+ x ≤ x,

T(ρ1, ψ) < A(l)

[

T(ρ, f ) +

(

1ρ − ρ1

)

+ ρ +O(1)

]

that is, log+ T(ρ1, ψ) < log+ T(ρ, f ) + log+1

ρ − ρ1+ log+ ρ +O(1).74

Substituting this and remembering thatr < ρ1 < ρ, ρ−ρ1 = ρ1− r =12(ρ − r) we get the result. �

Now we are ready to prove theorem 13.


Proof of theorem 2.13.We have from theorem 8,

m(r, ψ) +m

(

r,1ψ

)

+m

(

r,1

ψ − 1

)

≤ 2T(r, ψ) − N1(r, ψ) + S(r, ψ)

i.e.

T

(

r,1ψ

)

+ T

(

r,1

ψ − 1

)

≤ T(r, ψ) − N1(r, ψ) + S(r, ψ)

+ N(r, ψ) + N

(

r,1ψ

)

+ N

(

r,1

ψ − 1

)

i.e.

T(r, ψ) ≤ N(r, ψ) − N1(r, ψ) + N

(

r,1ψ

)

+ N

(

r,1

ψ − 1

)

+ S(r, ψ) +O(1)

also

N(r, ψ) − N1(r, ψ) = N(r, ψ) − N

(

r,1ψ′

)

and

N

(

r,1

ψ − 1

)

− N

(

r,1ψ′

)

= N

(

r,1

ψ − 1

)

− N0

(

r,1ψ′

)

. Hence

Thus,

T(r, ψ)N(r, ) + N(r, 1/ ) + N

(

r,1−1

)

− N0(r, 1/1)+ S(r, ) +O(1)

where N0(r, 1/ψ′) denotes the fact that zeros ofψ′ at multiple rootsof ψ − 1 are omitted. Thus, sinceT(r, ψ) = m(r, 1/ψ) + N(r, 1/ψ) +

O(1)m(r, 1/ψ) ≤ N(r, ψ) + N

(

r,1

ψ − 1

)

− N0(r, 1/ψ′) + S(r, ψ) + O(1).

Note again that poles ofψ occur only at poles off or of aν(z), and inN 75

each pole is counted only once. Then

N(r, ψ) ≤ N(r, f ) +∑

ν

N(r, aν)


≤ N(r, f ) +OT(r, f ).

Again,

T(r, f ) = m(r, 1/ f ) + N(r, 1/ f ) +O(1)

≤ m(r, ψ/ f ) +m(r, 1/ψ) + N(r, 1/ f ) +O(1).

Substituting we obtain,

[O(1)+ 1]T(r, f ) ≤ N(r, f ) + N

(

r,1

ψ − 1

)

+ N(r, 1/ f ) − N0(r, 1/ψ′)

+S(r, ψ) +m(r, ψ/ f ),

since f (z) being admissibleO(1) = O[T(r)]. Now if 0 < r < ρ < Rlemmas 6 and 8 give

m(r, ψ/ f ) + S(r, ψ) < A(l)

[


ρ − r+ log+ ρ +O(1)

]

.

and now the result is completed just as in theorem 9 by means oflemma4. Hence theorem 13.

Consequences.

Theorem 14(Milloux) . If f (z) is admissible in|z| < R and is regularthere then either f(z) assumes every finite value infinitely often or ev-ery derivative of f(z) assumes every finite value except possibly zero,infinitely often.

Proof. Supposef (z) = a, f (l)(z) = b have only a finite number of roots

whereb , 0. Chooseg(z) = f (z)−a, ψ(z) =g(l)(z)

b=

f (l)(z)b

in theorem76

13. Then sinceN(r, g) = 0, g being regular

[1 + 0(1)]T(r, g) < N

[

r,1

g(l)(z) − b

]

+ N(r, 1/g) + S1(r)

where limr→R

S1(r)/T(r, g) = 0.


If R = ∞ this gives limr→∞

T(r, g)/ log r < +∞ by assumptions that

n

[

t,1

f (l)(z) − b

]

andn(t, 1/g) are finite asr tends to infinity so thatg(z)

is rational and so isf (z). That is f (z) is a polynomial. �

If R is less than infinity we obtain, limr→R

T(r, g) < ∞, and so since

T(r, g) increases withr lim T(r, g) < ∞ and the same result applies tof (z) giving a contradiction to admissibility.

Theorem 15(Saxer). If f (z) is meromorphic in the plane and f , f′, f ′′

have only a finite number of zeros and poles then f(z) = P1(z)/P2(z)eP3(z) P1, P2, P3 being polynomials. If f , f′, f ′′ have no zeros and polesthen f(z) = ea+bz where a, b are constants.

Proof. Setg(z) = f (z)/ f ′(z). Theng′(z) = 1− [ f (z) f ′′(z)/ f ′2(z)]. Sup-pose thatg(z) is transcendental and so admissible in the plane. Nowg = 0,∞ only when f ′ = 0, on f = 0,∞ that is a finite number of timesand so,N(r, g) + N(r, 1/g) = O(log r). Next g′(z) = 1 only for f = 0

or f ′′ = 0 that isN

(

r,1

g′ − 1

)

= O(log r) by hypothesis. Now theorem

13 applied tog(z) gives for a sequence ofr tending to infinity, takingψ = g′,

[1 +O(1)]T(r, g) = O(log r).

This impliesg(z) is rational, i.e. a contradiction. Henceg(z) is rational 77

so that f ′/ f = g is rational. Now f ′/ f has simple poles with integerresidues at the poles and zeros off (z). Since f ′/ f is rational by expand-ing it in partial fractions we get,

f ′/ f =∑

r

kr/(z− zr ) + P(z)

with kν integers andP(z) a polynomial. Integrating the above,

f (z) =∏

r

(z− zr)k∫

e

P(z)dz

This proves the first part and iff (z) has no zeros or poles the productterm disappears andf = eP(z), f ′(z) = P′(z)eP(z) and f ′(z) , 0 impliesP′(z) , 0, which givesP(z) = a+ bz. �


Remark . Note that we cannot do the same thing withf (z) and f ′(z).For if f (z) = eg(z), f ′(z) = g′(z)eg(z) and if we putg′(z) = eh(z) whereh(z) is an arbitrary integral function, so thatg(z) =

∫

eh(z)dzand f (z) =

exp·[∫

eh(z)dz]

. Then f is an integral function withf , 0, and f ′ , 0.So in theorem 15 we cannot leave out the restriction onf ′′. FurtherF(z) =

∫

f (z)dz is a function for whichF′ , 0, F′′ , 0 and so wecannot leave the restriction onf .

But we will show that we can leave out the restriction onf ′. This isprecisely Theorem 19.

In connection with theorem 15 we also quote the following exten-78

sion by Csillag [1].

Theorem 16. If l and m are different positive integers and f(z) an inte-gral function such that f(z) , 0, f (l)(z) , 0 and f(m)(z) , 0, then f(z) isequal to eaz+b.

2.6 Elimination of N(r, f )

We shall prove the following theorem

Theorem 17. If f (z) is admissible in|z| < R, and l≥ 1 then,

T(r, f ) < [2 + (1/l)]N(r, 1/ f ) + 2[1+ (1/l)]N

(

r,1

f (l) − 1

)

+ S2(r, f ),

wherelimr→R

S2(r, f )/T(r) = 0.

Let us setψ(z) = f (l)(z) in theorem 13 (Th. of Milloux), to get(2.7)

T(r, f ) < N(r, f ) + N(r, 1/ f ) + N

(

r,1

f (l) − 1

)

− N0

(

r,1

f (l+1)

)

+ S1(r, f )

where inN0

(

r,1

f (l+1)

)

zeros of f (l+1) at multiple roots off (l)(z) = 1 are

to be omitted.Further we need,


Lemma 9. If g(z) =[

f (l+1)(z)]l+1

/[1 − f (l)(z)] l+2 then

lN1(r, f ) ≤ N2(r, f ) + N

(

r,1

f (l) − 1

)

+ N0

(

r,1

f (l+1)

)

+m(r, g′/g)

+ log |g(0)/g′(0)|(2.8)

where N1(r, f ) stands for the N-sum over the simple poles of f(z); 79

N2(r, f ) for the N sum over multiple poles of f(z), with each pole countedonly once. ThusN(r, f ) = N1(r, f ) + N2(r, f ).

Now at a simple polez0 of f (z), f (z) = O(1)+ [a/(z− z0)] whereais not equal to zero.

Differentiatingl times,

1− f (l)(z) =al!(−1)l+1

(z− z0)l+1+O(1)

This can be written as

1− f (l)(z) =al!(−1)l+1

(z− z0)l+1[1 +O{(z− z0)l+1}]

The differentiation of both the sides again gives,

f (l+1)(z) =[

1+O{(z− z0)l+2}] a(l + 1)!(−1)l+1

(z− z0)l + 2

Hence,

g(z) =(−1)l+1(l + 1)l+1

a(l)![1 +O{(z− z0)l+1}]

So,g(z0) , 0,∞ but g′(z) has a zero of order at leastl at z = z0. Nowwe have

N(r, g/g′) − N(r, g′/g) = m(r, g′/g) −m(r, g/g′) + log |g(0)/g′(0)|

from Jensen’s formula. As we saw at the end of section 2 the left hand 80

side is

N(r, g) + N(r, 1/g′) − N(r, g′) − N(r, 1/g)


= N(r, 1/g′) − N(r, 1/g) − N(r, g)

= N0(r, 1/g′) − N(r, 1/g) − N(r, g)

where inN0(r, 1/g′) only zeros ofg′ which are not zeros ofg are to beconsidered. By our above analysis,

lN1(r, f ) ≤ N0(r, 1/g′)

N(r, 1/g) + N(r, g) +m(r, g′/g) + log |g(0)/g′(0)|

Then note thatg = 0,∞ only at poles off (z) which must be multiple, atzeros off (l)(z)−1, and zeros off (l+1)(z) which are not zeros off (l)(z)−1.This gives lemma 9.

Now (2.7) gives on writingT(r, f ) = m(r, f ) + N(r, f )(2.9)

N(r, f ) − N(r, f ) < N(r, 1/ f ) + N

(

r,1

f (l+1)

)

− N0

(

r,1

f (l+1)

)

+ S1(r, f )

On the left the contribution of each multiple pole to the sum beingcounted once forN, but at least twice forN. So,

(2.10) N2(r, f ) ≤ N(r, f ) − N(r, f )

Also N(r, f ) = N2(r, f ) + N1(r, f ). Hence it follows from lemma 9 that,

N(r, f ) ≤ [1 + (1/l)]N2(r, f )

+1l

[

N

(

r,1

f (l) − 1

)

+ N0

(

r,1

f (l+1)

)

+m(r, g′/g)

]

+1l

log|g(0)||g′(0)|

By the inequalities (2.9) and (2.10) it is at the most,81

(

1+1l

) [

N(r, 1/ f ) + N(r, 1/( f (l) − 1))− N0

(

r,1

f (l+1)

)

+ S1(r, f )

]

+ (1/l)

[

N

(

r,1

f (l) − 1

)

+ N0

(

r,1

f (l+1)

)

+m(r, g′/g) + log |g(0)/g′(0)|

]

= (1+ 1/l)N(r, 1/ f ) + (1+ 2/l)N(r, 1/( f (l) − 1))

− N0

(

r,1

f (l+1)

)

+ S3(r)


where

S3(r) = [1 + (1/l)]S1(r, f ) + (1/l)[m(r, g′/g) + log |g(0)/g′(0)|].

Substituting forN(r, f ) in (2.7) we obtain,

T(r, f ) < [2 + (1/l)]N(r, 1/ f ) + 2[1+ (1/l)]

N

(

r,1

f (l) − 1

)

− 2N0

(

r,1

f (l+1)

)

+ S1(r, f ) + S3(r, f )

We observe that the above gives the result of theorem 17, provided wepose

S2(r, f ) = S1(r, f ) + S3(r, f )

= [2 + (1/l)]S1(r, f ) + (1/l)[m(r, g′/g) + log |g(0)/g′(0)|]

and in order to complete the proof it is sufficient to prove that ifr < ρ <R, andr tends toR,

m(r, g′/g) < A[log+ T(ρ, f ) + log+ 1/(ρ − r) + log+ ρ + 0(1)]

The above inequality is true for,

logg(z) = (l + 1) log f (l+1)(z) − (l + 2) log[1− f (l)(z)]

g′/g = (l + 1)[

f (l+2)/ f (l+1)]

+ (l + 2) f (l+1)/(1− f (l))

m(r, g′/g) ≤ m(r, f (l+2)/ f (l+1)) +m(

r, f (l+1)/( f (l) − 1))

+O(1)

Now the result follows from Lemma 8, withψ = f (l)−1 or f (l+1). Hence 82

the proof of theorem 17 is complete.

2.7 Consequences

Theorem 18. The result of theorem 14 extends to meromorphic func-tions without any further hypothesis. To be precise, if f(z) is admissiblein |z| < R and is meromorphic there, then either f(z) assumes every fi-nite value infinitely often or every derivative of f(z) assumes every finitevalue except possibly zero infinitely often.


The proof is as before by the consideration ofg(z) = [ f (z) − a]/binstead off (z); if the equationsf (z) = a and f (l)

= b have only a finitenumber of roots. For theng = 0 andg(l)

= 1 have only finite number ofroots and we can use Theorem 17.

Theorem 19. Theorem 15 (Saxer’s theorem) still holds if the hypothesisof f ′(z) is omitted. That is, if f(z) is meromorphic in the plane and f ,and f′′ have only a finite number of zeros and poles then,

f (z) = P1(z)/P2(z)eP3(z)

P1, P2, P3 being polynomials. If f′′, f have no zeros and poles thenf (z) = eaz+b, a and b being constants.

Proof. If f (z) and f ′′(z) have only a finite number of zeros and poles,83

put g(z) = f (z)/ f ′(z). Theng′(z) = 1 − { f (z) f ′′/[ f ′(z)]2}g(z) has onlyfinite number of zeros and so doesg′(z) − 1 = f f ′′/ f ′2 namely at thepoles of f and at the zeros off and f ′′. Hence by Theorem 18,g(z)is rational and now the proof is completed as in Saxer’s theorem. Iff (z) and f ′′(z) have no zeros or poles,f (z) = eP(z) where P(z) is apolynomial. Therefore,f ′′(z) equals [P′′(z) + P′(z)2]eP(z), and if P′(z)has degreen greater than or equal to 1, thenP′(z)2, P′′(z) have degree2n, n − 1, respectively, so thatf ′′(z) has 2n zeros. ThusP′(z) = a =const.P(z) = az+ b. �

A slightly more delicate analysis shows that iff (z), f ′′(z) have nozeros butf (z) may have a finite number of poles, thenf (z) = eaz+b or(az+ b)−n, wheren is a positive integer.

Part III

Univalent Functions

3.1 Schlicht functions84

Definition. A function f(z) regular in a domain D is said to beunivalent(Schlicht, Simple) if f(z) takes different values at different points of D.Then f(z) maps D, one-one conformally into a domain∆.

Since f (z) takes no value more than once, iff (z) is Schlicht in theplane, the functionA(r) (area on the Riemann sphere) is less or equalto one andT0(r) ≤ log r. So f (z) is rational and is in fact a polynomialwhich must be linear. We consider functionsf (z) univalent in|z| < 1.

If f (z) =∞∑

n=0anzn, ( f (z) − a0)/a1 is also univalent. In facta1 cannot be

zero, for otherwisef (z) would assume values at least twice nearz = 0,a1 , 0. Hence we may assumef (z) to be of the form

(2) z+∞∑

n=2

anzn

The class of functions, univalent in|z| < 1 with the expansion (2) iscalledS.

The first two results are,

Theorem 1. If f (z) ∈ S , |a2| ≤ 2, and equality is possible only for

f (z) = fθ(z) =z

[1 − zeiθ]2, and

73

74 Univalent Functions

Theorem 2. If f (z) ∈ S and w is a value not taken by f(z) then|w| ≥ 14,

again equality is possible only for f(z) = fθ(z).

Both results in the above form are due to Biberbach, theorem 2with85

a smaller constant than14 is due to Koebe though it seems we had theproof eventually form

Note that f (z) ∈ S, takes all values with modulus< 14. We shall

prove theorem 1 first and then deduce theorem 2 from it. For that weneed

Lemma 1. Suppose f(z) is regular and univalent in the annulus r1 <

|z| < r2 then the area of the image of the annulus isπ∞∑

−∞n|an|

2(r2n2 − r2n

1 ),

where f(z) has the expansion∞∑

−∞anzn in the annulus.

Proof. The area of the image is clearlyr2∫

r1

r dr2π∫

0

| f ′(reiθ)|2dθ. The inte-

gral

2π∫

0

| f ′(reiθ)|2dθ =

2π∫

0

f ′(reiθ) f (reiθ)dθ

=

2π∫

0

[∑

nanrn−1ei(n−1)θ] [

∑

mamrm−1e−i(m−1)θ]

Since the multiplication of the two series and then term by term integra-

tions are valid, and2π∫

0

ei(m−n)θdθ =

0, m, n,

2π, m= n.

2π∫

0

| f ′(reiθ)|2dθ = 2π∞∑

−∞

n2|an|2r2n−2

∴ Area of image86

Univalent Functions 75

=

r2∫

r1

r dr

2π∫

0

| f (reiθ)|2dθ

= 2π

r2∫

r1

+∞∑

−∞

n2|an|2r2n−1dt.

Again since integration term by term is valid,

= π

+∞∑

−∞

n|an|2(r2n

2 − r2n1 )

Hence the lemma. �

We now proceed to prove the theorem.Consider the functionf (z) ∈ S.

f (z) = z+∞∑

2

anzn, let F(z) =[

f (z2)]

12= z

[

f (z2)

z2

]12

= z

1+∞∑

2

anz2n−2

12

Sincef (z)z

is not zerof (z)z

has a one valued square root which is a

power series inz and hence

[

f (z2)

z2

]12

is a power series inz2.

F(z) = z+12

a2z3+ · · · · · ·

F(z) is odd. AlsoF(z) is univalent. For ifF(z1) = F(z2), then

[

f (z21)]

12=

[

f (z22)]

12

i.e., 87

f (z21) = f (z2

2) =⇒ z21 = z2

2.


∴ z1 = ±z2.But F(−z1) = −F(z1) , F(z1) unlessz1 = 0. Soz1 = z2. Now set

g(z) =1

F(z)=

[

f (z2)]− 1

2=

12+ b1z+ b3z3

+ · · · · · ·

=1z+

∞∑

1

bnzn

whereb1 = −12a2.

And g(z) is univalent in 0< |z| < 1. Let J(r) be the curve which isthe image of|z| = r by g(z), 0 < r < 1 andA(r) the area inside it. Thenfor 0 < r1 < r2 < 1 we have by the lemma

A(r1) − A(r2) = ±π

1

r21

−1

r22

+

∞∑

1

n|bn|2(r2n

2 − r2n1 )

SinceJ(r1) andJ(r2) do not cross, the left hand side and hence the righthand side is different from zero. SoA(r) is monotonic. Further for smallr, A(r) ∼

π

r2which tends to∞ asr → 0. HenceA(r) decreases. The

left hand side is therefore positive and the quantity insidethe brackets ispositive and so we take the positive sign.

SetS(r) =1

r2−∞∑

1|bn|

2nr2n.

ThenA(r) = πS(r) + C, C being a constant. We want to prove that88

S(r) ≥ 0 for 0< r < 1.Suppose now thatb1 i.e. a2 is real Otherwise ifa2 = |a2|eiθ we con-

sidereiθ f (ze−iθ) in place of f (z) and then

eiθ f (ze−iθ) = z+ eiθa2e−2iθz2+ · · · · · ·

= z+ |a2|z2+ · · · · · ·

Thus there is no loss of generality in assuminga2 real and positive. Now

g(z) =1z+ b1z+ b3z3

+ · · · · · · , so if, z= reiθ

g(reiθ) =

(

1r+ b1r

)

cosθ + i

(

b1r −1r

)

sinθ +O(r3).


We have then

|Rl. g(reiθ)| ≤∣

∣

∣

∣

∣

1r+ b1r

∣

∣

∣

∣

∣

+O(r3)

| Im .g(reiθ)| ≤∣

∣

∣

∣

∣

1r− b1r |

∣

∣

∣

∣

∣

+O(r3) =

(

1r− b1r

)

+O(r3)

(sincea2 > 0, b1 < 0) so that the image of|z| = r by g(z) is contained

in the ellipse of semi-axes

(

1r− b1r

)

+ O(r3), |1r+ b1r | +O(r3). Hence

area inside the imageA(r) satisfies

A(r) ≤ π

(

1

r2− b2

1r2)

+O(r2) =π

r2+O(r2).

Thus 89

πS(r) +C ≤π

r2+O(r2).

But

S(r) =1

r2+O(r2)

so that1

r2+O(r)2

+C ≤π

r2+O(r2)

orC +O(r2) ≤ O(r2)

letting r → 0 we see thatC ≤ 0. This proves that−C+A(r) = πS(r) ≥ 0since,−C ≥ 0, A(r) ≥ 0. Thus for 0< r < 1 S(r) ≥ 0. [Actually a littlemore refined argument shows thatA(r) = πS(r)].

Therefore,1

r2≥

∞∑

1

|bn|2nr2n

and lettingr → 1, 1≥∞∑

1n|bn|

2. Thus

|b1| ≤ 1 or |a2| ≤ 2


Equality can hold only ifbn = 0 for n > 1 i.e.g(z) =1z− zeiθ.

i.e. F(z) =z

1− z2eiθ= [ f (z2)]

12

Thusf (z) =

z

[1 − zeiθ]2= fθ(z).

This proves the theorem provided we showfθ(z) ∈ S. To deduce theo-

rem 2, setg(z) =w f(z)

w− f (z)where f (z) , w for |z| < 1.

Then

g(z) =w(z+ a2z2

+ · · · · · · )w− (z+ a2z2 + · · · .)

= (z+ a2z2+ · · · )

(

1+zw+

a2z2

w+ · · ·

)

by expansion90

= z+

(

a2 +1w

)

z2+ higer powers ofz.

Since the map is bi-linear,g(z) is also univalent. In factg(z1) = g(z2)implies, f (z1) = f (z2) from which it follows thatz1 = z2. Furtherg(z) ∈ S. Hence by the above theorem,

∣

∣

∣

∣

∣

∣

(

1w+ a2

)∣

∣

∣

∣

∣

∣

≤ 2∣

∣

∣

∣

∣

1w

∣

∣

∣

∣

∣

≤ 2+ |a2|

≤ 4

∴ |w| ≥14

and equality is possible only if|a2| = 2, i.e. f (z) = fθ(z).

Note that

f0(z) =z

(1− z)2=

14

[

(1+ z)2

(1− z)2− 1

]

if ζ =1+ z1− z

then by this linear transformation|z| < 1 corresponds to

real ζ > 0 i.e.,Z = ζ2 gives the plane cut along the negative axis. So


f0(z) is univalent and in factf0(z) maps onto the plane cut from−14

to ∞ along the real axis. Hencef0(z) is univalent possessing such anexpansion defined for elements ofS and so doesfθ(z) = e−iθ f0(zeiθ).

Also fθ(z) , −14

e−i becausef0(z) , −14

. Note also that

fθ(z) = z+∞∑

2

nznei(n−1)θ

so that|an| = n for all n.

Theorem 3. Suppose f(z) ∈ S and|z| = r, 0 < r < 1 then the following 91

inequalities hold.

r

(1+ r)2≤ | f (z)| ≤

r

(1− r)2

1− r

(1+ r)3≤ | f ′(z)| ≤

1+ r

(1− r)3

(1− r)r(1− r)

≤

∣

∣

∣

∣

∣

f ′(z)f (z)

∣

∣

∣

∣

∣

≤1+ r

r(1− r)

where equality is possible only for functions fθ(z) defined already.

Proof. Assume|z0| < 1 and set

ϕ(z) = f

[

z0 + z1+ z0z

]

= b0 + b1z+ · · · · · ·

Sincez0+ z1+ z0z

= w is a bi-linear map of the unit circle onto itself.ϕ(z) is

univalent in|z| < 1 and soϕ(z) − b0

b1∈ S.

Applying theorem 1 we deduce,

∣

∣

∣

∣

∣

b2

b1

∣

∣

∣

∣

∣

≤ 2

|b2| ≤ 2|b1|


we have

ϕ′(z) = f ′(

z+ z0

1+ z0z

) [

11+ z0z

−(z0+ z)z0

(1+ z0z)2

]

= f ′[

z+ z0

1+ z0z

] {

1− (z0)2

(1+ zz0)2

}

and92

ϕ′′(z) = f ′′(

z+ z0

1+ z0z

) [

11+ z0z

−(z0 + z)z0

(1+ z0z)2

]2

− 2 f ′(

z+ z0

1+ z0z

) [

(1− |z0|2)z0

(1+ zz0)2

]

Thus

b1 = ϕ′(0) = (1− |z0|

2) f ′(z0)

b2 =12ϕ′′(0) =

12

[1 − |z0|2]2 f ′′(z0) − f ′(z0)z0(1− |z0|

2).

We have seen|b2| ≤ 2|b1| i.e.

(3.1) | f ′′(z0)(1− |O|2)2 − 2z0 f ′(z0)(1− |z0|2)| ≤ 4(1− |z0|

2)| f ′(z0)|

If z0 = ρeiθ this gives∣

∣

∣

∣

∣

∣

f ′′(z0)f ′(z0)

z0 −2ρ2

1− ρ2

∣

∣

∣

∣

∣

∣

≤4ρ

1− ρ2, ρ < 1

Now for any complex function∂w∂r=

dwdz

eiθ. If z = reiθ, and so

∂

∂r[log f ′(z)] =

f ′′

f ′eiθ. Thus the above inequality is

∣

∣

∣

∣

∣

∣

ρ∂

∂ρlog f ′(z) −

2ρ2

1− ρ2

∣

∣

∣

∣

∣

∣

≤4ρ

1− ρ2, (z0e−iθ

= ρ),

i.e.2ρ − 4

1− ρ2≤

∂

∂ρ[log | f ′(ρeiθ)] ≤

2ρ + 4

1− ρ2


because log| f ′(z)| = Rl. log f ′(z).Now integrate the above with regard toρ,

log1

1− ρ2− 2 log

11− ρ

− 2 log(1+ ρ) ≤ log | f ′(ρeiθ)|

≤ log1

1− ρ2+ 2 log

11− ρ

+ 2 log(1+ ρ).

1− ρ

(1+ ρ)3≤ | f ′(z)| ≤

1+ ρ

(1− ρ)3, z= ρeiθ .

This gives bounds forf ′(z). 93

Again,

| f (reiθ)| =

∣

∣

∣

∣

∣

∣

∣

∣

∣

r∫

0

f ′(ρeiθ)dρ

∣

∣

∣

∣

∣

∣

∣

∣

∣

≤

r∫

0

| f ′(ρeiθ)|dρ ≤

r∫

0

1+ ρ

(1− ρ)3dρ =

r

(1− r)2

To get the lower bound forw = f (ρeiθ), we suppose|w| < 14, since for

|w| ≥ 14 the result is trivial

(

ρ

(1+ρ)2 <14

)

. Thus by theorem 2 (Section 3)

the line segment [0,w] lies entirely in the image of|z| < 1 by w = f (z).If λ is this line segment,γ the corresponding curve in thez-plane

|w| =∫

λ

|dw| =∫

γ

∣

∣

∣

∣

∣

dwdz

∣

∣

∣

∣

∣

|dz| ≥

ρ∫

0

1− ρ(1+ ρ)3

dρ.

=ρ

(θ + ρ)2

because∣

∣

∣

∣

∣

dwdz

∣

∣

∣

∣

∣

|dz| ≥1− ρ

(1+ ρ)3dρ if dρ is positive sincedρ < |dz|, if dρ < 0

the result is even more evident.

Sinceϕ(z) − b0

b1∈ S, we obtain, 94

ρ

(1+ ρ)2≤

∣

∣

∣

∣

∣

∣

ϕ(ρeiθ) − b0

b1

∣

∣

∣

∣

∣

∣

≤ρ

(1− ρ)2.

Now z0 = ρeiθ.


Thenϕ(z0) = 0, b0 = f (z0), b1 = [1 − |z0|2] f ′(z0)

|z0|

1+ |z0|2≤| f (z0)|

[1 − |z0|2]| f ′(z0)| ≤

|z0|

1− |z0|2

i.e.r(1− r)1+ r

≤

∣

∣

∣

∣

∣

f (z)f ′(z)

∣

∣

∣

∣

∣

≤r(1+ r)1− r

.

�

This gives the bounds for∣

∣

∣

∣

∣

f ′

f

∣

∣

∣

∣

∣

since equality holds in (3.1) only for

ϕ(z) = fθ(z) it can be shown that for all the inequalities of theorem 3equality is possible only for this function.

Theorem 4(Littlewood, Paley, Spencer). Suppose f(z) ∈ S and for anyλ→ 0 set

(3.2) Iλ(r, f ) =12π

π∫

−π

| f (reiθ)|λdθ

Sλ(r) = rddr

Iλ(r).

Then

(3.3) Sλ(r) =λ2

2π

r∫

0

ρdρ

π∫

−π

| f (ρeiθ)|λ−2| f ′(ρeiθ)|2dθ

Thus

Sλ(r) ≤ λM(r, f )λ ≤λrλ

(1− r)2λ,(3.4)

Iλ(r) =

r∫

0

Sλ(ρ)dρρ

≤ λ

r∫

0

M(ρ, f )λdρρ

≤

A(λ)[1 − r]1−2λ, λ > 12

A log1

1− rλ = 1

2

A(λ) λ <12

(3.5)

95


Proof. Suppose

f (z) = z

1+∞∑

2

anzn−1

Set

ϕ(z) = [ f (z)]λ/2 = z12λ

1+∞∑

1

bnzn

This is possible sincef (z)z

is regular and non-zero in|z| < 1 and

so has aλ

2th power which is also regular. For definiteness we take the

principal value ofzλ/2, |argz| < π.

We haveϕ′(z) =∞∑

0

[

n+λ

2

]

bnzn+ 12λ−1.

12π

π∫

−π

|ϕ′(ρeiθ)|2dθ =12π

π∫

−π

ϕ′ϕ′dθ =∞∑

n=0

[

n+λ

2

]2

|bn|2ρ2n+λ−2

exactly as in lemma 1 (Section 3)

Iλ(r, f ) =12π

π∫

−π

|ϕ(reiθ)|2dθ =12π

π∫

−π

ϕϕdθ =∞∑

0

|bn|2ρ2n+λ

Sλ(r, f ) = rddr

Iλ(r, f ) =∑

(2n+ λ)|bn|2r2n+λ.

Further 96

r∫

0

ρdρ

π∫

−π

|ϕ′(ρeiθ)dθ =π

2

∑

(2n+ λ)|bn|2r2n+λ

=π

2Sλ(r, f ).

Now ϕ′ =λ

2f ′ f

λ−12

|ϕ′|2 =λ2

4| f ′|2| f λ−2|


Thus

Sλ(r, f ) =λ2

2π

r∫

0

ρdρ

π∫

−π

| f ′(ρeiθ)|2| f (ρeiθ)|λ−2dθ.

giving (3.3).We now interpret the above formula.ρdρ dθ is an element of area

in |z| < 1, | f ′|2ρdρdθ is the corresponding area in the image plane atw = Reiφ and | f |λ−2| f ′|2ρdρ dθ the mass of the area, if we imagine amass densityRλ−2 on the circle|w| = R in the image plane. Then ourintegral is the total mass of the image. Since the image covers no pointmore than once and lies in|w| < M(r, f ) = M say, the mass of image≤total mass of the circle which is

M∫

0

2πR · Rλ−2dR=2πλ

Mλ and so

Sλ(r) ≤λ2

2π2πλ

Mλ= λMλ ≤ λ

rλ

(1− r)2λ

This gives (3.4), because from theorem 3,

M(r, f ) ≤r

(1− r)2.

Thus97

Iλ(r) ≤ λ

r∫

0

ρλ−1

(1− theta)2λdρ. If λ ≥ 1, ρλ−1 ≤ 1 and so

Iλ(r) ≤ λ

r∫

0

1

(1− ρ)2λdρ ≤

λ

2λ − 1(1− r)1−2λ.

Assume thatλ < 1. Then

Iλ(r) ≤ λ

r∫

0

ρλ−1

(1− ρ)2dρ =

ρλ

(1− ρ)2λ

r

0

− 2λ

r∫

0

ρλ

(1− ρ)2λ+1dρ.


≤rλ

(1− r)2λ− 2λ

r∫

0

ρ

(1− ρ)2λ+1dρ since ρ ≤ ρλ if 0 ≤ λ ≤ 1

=rλ

(1− r)2λ+ 2λ

r∫

0

1

(1− ρ)2λdρ − 2λ

r∫

0

1

(1− ρ)2λ+1dρ

=rλ

(1− r)2λ+

2λ2λ − 1

(1− r)1−2λ −2λ

2λ − 1−

1

(1− r)2λ+ 1 if 2λ , 1.

≤2λ

2λ − 1(1− r)1−2λ −

2λ2λ − 1

+ 1 ≤2λ

2λ − 1(1− r)1−2λ

Hence if 1− 2λ > 0, (1− r)1−2λ ≤ 1, Iλ(r) ≤2λ

2λ − 1.

If 2λ = 1, Iλ(r) ≤ 12

r∫

0

1

ρ12 (1−ρ)

dρ =r∫

0

dρ

1− ρ298

=12

log1+ r1− r

=12

log1− r2

(1− r)2≤

12

log1

(1− r)2= log

1(1− r)

Thus

Iλ(r) ≤

2λ2λ − 1

(1− r)1−2λ if λ >12

log1

1− rif λ =

12

2λ2λ − 1

if λ <12

This proves the theorem completely. �

Theorem 5 (Littlewood). If f (z) = z+∞∑

2anzn ∈ S then|an| < e for

n ≥ 2.

Proof. We have|an| =

∣

∣

∣

∣

∣

∣

∣

12π

∫

|z|=ρ

f (z)dzzn+1

∣

∣

∣

∣

∣

∣

∣

≤1

2πρn

π∫

−π

| f (ρeiθ)|dθ =I1(ρ, f )ρn


By inequality (3.5)

I1(ρ, f ) ≤

ρ∫

0

dt

(1− t)2=

ρ

1− ρ.

Thus |an| <1

ρn−1(1− ρ). We chooseρ so that

1

ρn−1(1− ρ)is minimal

i.e. putρ = 1− =n− 1

nto get|an| <

( nn− 1

)n−1n

i.e., |an| <

(

1+1

n− 1

)n−1

n < en

This completes theorem 5. �99

Remark. Bazilevic [1] improved this toI1(ρ, f ) <ρ

1− ρ2+ 0.55 so that

we get

|an| <12

en+ 1.51

Theorem 6. Suppose f(z) ∈ S and set

ϕ(z) = [ f (z)]λ ϕ(z) = zλ

∞∑

0

an, λzn

.

Then ifλ > 14, |an,λ| < A(λ)n2λ−1. In particular if f (z) = z+ ak+1zk+1

+

· · ·+akn+1zkn+1+ · · · ∈ S , then|a2n+1| < A1 if k = 2 and|a3n+1| < A2n−1/3

if k = 3, A1 and A2 are absolute constants.

Proof. Now (n+ λ)|an,λ | =

∣

∣

∣

∣

∣

∣

∣

12πi

∫

|z|=ρ

ϕ′(z)dz

zn+λ

∣

∣

∣

∣

∣

∣

∣

(3.6) ≤1

ρn+λ−1I1(ρ, ϕ′) . . .

with the notation of theorem 4. �


Sinceϕ′ = λ f λ−2 f ′

I1(ρ, ϕ′) =λ

2π

2π∫

0

| f ′(ρeiθ)|| f (ρeiθ)|λ−1dθ

=λ

2π

2π∫

0

| f ′(ρeiθ)|| f (ρeiθ)|t−1| f (ρeiθ)|λ−tdθ

for any t

I1(ρ, ϕ′) ≤

12π

2π∫

0

| f ′(ρeiθ)|2| f (ρeiθ)|2t−2dθ

12

×

λ

12π

2π∫

0

| f (ρeiθ)|2λ−2tdθ

12

(3.7)

by Schwarz inequality. 100

Sinceλ >14

, we chooset sufficiently small but positive such that

2λ − 2t > 12; for examplet = 1

2(λ −14

) can be a choice oft. By theorem

4

1− 12n

∫

1− 1n

ρdρ

2π∫

0

| f ′(ρeiθ)|2| f (ρeiθ)|2t−2dθ ≤π

2t2S2t

(

1−12n, f

)

≤ A(t)n4t

Suppose

2πµ = min.1− 1

n≤ρ≤1− 12n

π∫

0

| f ′(ρeiθ)|2| f (ρeiθ)|2t−2dθ


Hence

(3.8) µ ≤A(t)n4t

1− 12n

∫

1− 1n

ρ dρ

≤ A(t)n4t+1

Also by theorem 4,

2π∫

0

| f (ρeiθ)|2λ−2tdθ ≤ A(λ, t)(1− ρ)−4λ+4t+1 since 2λ − 2t ≥12

≤ A(λ)n4λ−4t−1(3.9)

sincet depends uponλ.101

Hence for this valueρ from (3.7),

I1(ρ, ϕ′) ≤ A(λ)n(4t+1+4λ−4t−1)12 = A(λ)n2λ

Therefore from (3.6) it follows

(n+ λ)|an,λ | ≤A(λ)n2λ

ρn+λ−1<

(

1−1n

)−n−λ

A(λ)n2λ

≤ A(λ)n2λ

A(λ) just standing for a constant depending uponλ. Hence the result|an,λ| < A(λ)n2λ−1 follows.

Suppose now thatg(z) = z +∞∑

n=1akn+1zkn+1 ∈ S. Then so does

f (z) = [g(z1/k)]k= z

(

1+∞∑

n=1akn+1zn

)k

. For clearly f (z) is regular in

|z| < 1. f (0) = 0, f ′(0) = 1. Supposef (z1) = f (z2). Then

[

g(

z1k1

)]k=

[

g(

z1k2

)]k=⇒ g

(

z1k1

)

= ωg(

z1k2

)

= g(

z1k2ω

)

whereω is akth root of unity. Butg(z) ∈ S, and we havez1k1 = ωz

1k2 =⇒

z1 = z2.


Thus

g(

z1k

)

= [ f (z)]1/k

= z1/k

1+∞∑

n=1

akn+1zn

Hence applying the first part, withλ = 1/k (providedk = 1, 2, 3) 102

|akn+1| < A(k)n as required

i.e. |an+1| < A1n if k = 1

|a2n+1| < A2 if k = 2

|a3n+1| < A3n−1/3 if k = 3

This inequality is false for largek as was shown by Little wood [1]even if f (z) is continuous in|z| ≤ 1. We do not know whether it is

true for anyk ≥ 4. If f (z) = z +∞∑

2anzn is bounded and univalent

then the area of the image of|z| < 1 is at mostπM2 whereM is theleast upper bound of| f (z)|. Hence

∑

n|an|2 ≤ M2 and it follows that

|an| = O(n−12 ) asn → ∞. Nothing stronger than this is known. For

further discussion see Hayman Chapter 3. This is the correctorder formean-valent functions and probably forf (z) ∈ S also. The best exampledue to Clunie (unpublished) gives|an| > n−13/14 for some largen, so that

there is a gap between12

and 13/14. It can be shown that in theorem 6

the conclusion that|an| = O(1) obtains for alln if |an| = O(1) for somesequence ofn with constant common difference. We do not know if thisconclusion is till true i.e. if|an| = O(1) for a sequencen = nk such thatnk+1 − nk < constant.

In this connection Biernacki [1] has shown that for everyf (z) ∈ S,

||an+1| − |an|| < A[log(n)]3/2 for n ≥ 2.

Hence ifan = 0 for a sequencen = nk, with nk+1 − nk < const. |an| = 103

O(logn)3/2 for the intermediate coefficients. It is still to be found outwhetherO(1) can replaceO(logn)3/2.


Examples for theorem 6.

f (z) =z

(1− z)2

fk(z) = [ f (zk)]1/k=

z

(1− zk)2/k=

∑

an,kzkn+1

where

an,k =

2k

(

2k + 1

)

. . . . . .(

2k + n− 1

)

1 · 2 . . . . . . n

=

Γ

(

n+ 2k

)

Γ(n+ 1)Γ(2/k)∼

n2k−1

Γ(2/k)

so that theorem 6 is best possible.

Theorem 7 (Rogosinski, Deudonne’, Szasz). If f (z) = z+∞∑

2anzn be-

longs to S and has real coefficients then|an| ≤ n.

Proof. Let f = u+ iv. We have sincean are realf (z) = f (z). z= x+ iy,y , 0 impliesv , 0 for otherwise ifv = 0 for z = x + iy, y , 0 thenf (z) = f (z) = u which contradicts uni valency. Further we assert that forzsuch thaty > 0v must have constant sign. For if instead we havev(z1)positive andv(z2) negativez1 andz2 having the imaginary part positive,104

v(z) would be zero somewhere on the line joiningz1 andz2, which isnot possible. Clearly by consideringf (z)/z for smallz v> 0 for z withIm z> 0. The proof of the theorem essentially depends on this nature ofthe sign ofv. �

Now v(reiθ) =∞∑

1anrn sinnθ sincean are real wherea1 = 1.

Hence we have on integration,π∫

0

v(reiθ) sinθn dθ =π

2rnan since the

other terms vanish.Also | sinnθ| ≤ nsinθ for 0 ≤ θ ≤ π.It is enough to show this inequality for 0≤ θ ≤

π

2since| sinn(π −

θ)| = | sinnθ|, sin(π − θ) = sinθ.


Sinθθ

decreases in 0< θ ≤π

2.

Hence ifnθ ≤π

2i.e. θ ≤

π

2n

sinnθnθ

≤sinθθ

which gives the inequality forθ ≤π

2n.

In particular when

nθ =π

2

sin π2

π2

≤sin π

2nπ2n

or sinπ

2n≥ 1

n.

Hence ifπ

2n≤ θ ≤

π

2. 105

| sinnθ| ≤ 1 ≤ nsinπ

2n≤ nsinθ

Hence follows the inequality| sinnθ| ≤ nsinθ 0 ≤ θ ≤ π.

|an| =

∣

∣

∣

∣

∣

∣

∣

∣

∣

2πrn

π∫

0

v(reiθ) sinnθ dθ

∣

∣

∣

∣

∣

∣

∣

∣

∣

Now making use of the fact thatv > 0 for the range ofθ

|an| ≤2πrn

π∫

0

v(reiθ)| sinθn|dθ

≤2πrn

π∫

0

v(reiθ)nsinθ dθ

=n

rn−1a1 =

n

rn−1

Let us maker → 1,|an| ≤ n as required.


The functionz

(1− z)2∈ S has real coefficients, and satisfiesan = n.

Thus our inequality is sharp.We next proceed to prove another special case of Bieberbach’s con-

jecture. For this we need the definition:

Definition . A domain D is said to be star-like w.r.t. the origin O if forany point P∈ D, OP lies entirely in D.

Theorem 8(Nevanlinna). If w = f (z) = z+∞∑

2anzn ∈ S and maps|z| < 1106

onto a domain D star-like with respect to w= 0 then|an| ≤ n. Equalityis possible only when f(z) = fθ(z) defined already.

Before proceeding with the proof of the theorem let us prove thefollowing lemma due to Borel.

Lemma 2. If ϕ(z) = 1 +∞∑

n=1bnzn satisfies Rlϕ(z) ≥ 0 for z ≥ 0 then

|bn| ≤ 2, equality holds only forϕ(z) =1+ z1− z

= 1+∞∑

12zn.

Considerψ(z) =ϕ(z) − 1ϕ(z) + 1

. By hypothesisϕ(z) has real part+ve

which clearly implies|ϕ(z) − 1| < |ϕ(z) + 1| and so|ϕ(z)| < 1 and furtherψ(0) = 0.

Hence now applying Schwarz’s lemma

|ψ′(0)| ≤ 1. Equality holds only ifψ(z) = zeiθ.

Near

z= 0 ψ(z) =

(

∞∑

1bnzn

)

2+∞∑

n=1bnzn

=b1

2z+ · · · · · ·

which gives that|b1| ≤ 2. In order to deduce the inequality forbn,consider

ϕk(z) = 1+ bkz+ b2kz2+ · · · · · · + bnkz

n+ · · · · · ·


=1k

k∑

r=1

ϕ(wrz1/k)

wr running through thekth roots of unity and for a fixed value ofz1/k.Forwr = wr wherew = e

2πik . If we set

ζ = z1/k,1k

k∑

r=1

(wrz1/k) = 1+1k

∞∑

n=1

bnζn[wn

+ · · · + wkn]

= 1+∞∑

n=1

bnkζnk

= 1+∞∑

1

bnkzn

as wn+ · · · · · · + wkn

= 0 if k does not dividen

= k if k divides n.

Clearly Rlϕk(z) ≥ 0 sinceRlϕ(wrz1/k) ≥ 0 for every. Hence applying107

previous part toϕk(z), |bk| ≤ 2.

Proof of theorem 3.8.Note first that ifGr denotes the image of|z| < rby f (z) then ifG1 is star-like so isGr for 0 < r ≤ 1.

Considerζ = ϕ(z) = f −1[t f (z)] 0 < t < 1. Then sinceG1 is star-liket f (z) lies in the image of|z| < 1 by f (z) and so f −1[t f (z)] is a welldefined point in|ζ | < 1, and clearlyζ = 0 corresponds toz = 0. Thus|ϕ(z)| < 1 ϕ(0) = 0 and hence by Schwarz’s lemma

|ϕ(z)| ≤ |z|

t f (z) = f (ζ) where|ζ | ≤ |z| and so if|z| < r, f (ζ) ∈ Gr i.e. t f (z) ∈ Gr .This being true for anyt, 0< t < 1, Gr is star-like.

SinceGr is star-like its boundaryΓr meets any ray from 0 to∞ in 108

only one point. The limiting case whenΓr contains a line segment insuch a ray is excluded sinceΓr is a simple closed analytic curve. Thusarg. f (reiθ) increases withθ for any fixedr, 0< r < 1. Let

log f (reiθ) = u+ iv = log | f (reiθ)| + i arg. f (reiθ)


and arg. f (reiθ) is increasing for any fixedr implies∂v∂θ≥ 0.

i.e., Im.∂

∂θ[log f (reiθ)] ≥ 0

Im .

[

ireiθ f ′(reiθ)f (reiθ)

]

≥ 0

i.e., Rl.reiθ f ′(reiθ)

f (reiθ)≥ 0

In other words,

Rl.z f′(z)f (z)

≥ 0, 0 ≤ |z| < 1

Now applying the lemma withϕ(z) = z f′(z)/ f (z) and observing, thatthis function satisfies the hypothesis of the lemma, (forϕ(z) = 1+O(z)nearz= 0) Rl.ϕ(z) ≥ 0 and so if

ϕ(z) =z f′(z)f (z)

= 1+∞∑

1

bnzn

then|bn| ≤ 2.Again

z f′(z) =∞∑

1

nanzn= ϕ(z) f (z) =

∞∑

1

anzn

1+∞∑

n=1

bnzn

with a1 = 1.Equating coefficient ofzn on either side,109

nan = an + b1an−1 + b2an−2 + · · · · · · + bn−1a1

i.e. (n− 1)an = b1an−1 + b2an−2 + · · · · · · + bn−1a1

(n− 1)an ≤ 2[1+ |a2| + |a3| + · · · · · · + |an−1|] by the lemma.

Suppose we now assume that|aν| ≤ ν for ν ≤ n− 1 then we have,

(n− 1)|an| ≤ 2[1+ 2+ · · · · · · + n− 1]

≤ n(n− 1)


|an| ≤ n.

and by theorem 1,|a2| ≤ 2. Hence the proof is complete by induction onn. Further equality is possible forn ≥ 2 only if |a2| = 2 i.e. for functionsfθ(z) =

z(1−zeiθ)2 .

For extensions of this result to a more general class of imagedo-mains see Kaplan, W. [1].

3.2 Asymptotic behaviour

Theorem 9. If f (z) ∈ S and unless f(z) ≡ fθ(z) we have(1− r)2

rM(r, f )

decreasing steadily with increasing r and so tends toα where0 ≤ α < 1as r→ 1.

To prove this we require

Lemma 3. Suppose f(z) ∈ S and for fixedθ f (reiθ) = R(r)eiλ(r). Suppose 110

further that0 < r1 < r2 < 1 and r = r1, r2 correspond to R= R1, R2.Then

logR2(1− r2)2

r2≤ log

R1(1− r1)2

r1−

r1

4

r2∫

r1

(1− r)[λ′(r)]2dr.

Proof. We have

ddr

log(Reiλ) =ddr

log f (reiθ) = eiθ f ′(reiθ)f (reiθ)

.

On the other hand

ddr

log(Reiλ) =1R

dRdr+ iλ′(r)

By theorem 3∣

∣

∣

∣

∣

∣

f ′(reiθ)f (reiθ)

∣

∣

∣

∣

∣

∣

≤1+ r

r(1− r)so that


[

1r

dRdr

]2

+ [λ′(r)]2 ≤(1− r)2

r2(1− r)2,

saya2+ b2 ≤ c2 i.e. |a|2 + |b|2 ≤ c2.

This can be written as

(|c| − |a|)(|c| + |a|) ≤ b2

or |c| − |a| ≥b2

|c| + |a|≥

b2

2|c|

since|c| ≥ |a|. So|a| ≤ |c| −b2

2|c|. Sincea ≤ |a| we get

1R

dRdr≤

1− rr(1− r)

−r(1− r)2(1+ r)

[λ′(r)]2.

≤1+ r

r(1− r)−

r1

4(1− r)[λ′(r)]2

sincer1 ≤ r for the ranger1 ≤ r ≤ r2 and 1+ r < 2. Integrating fromr1111

to r2

logR2 − logR1 ≤

r2∫

r1

1+ rr(1− r)

dr −r1

4

r2∫

r1

(1− r)[λ′(r)]2.

=

r2∫

r1

1r(1− r)

dr +

r2∫

r1

dr −r1

4

r2∫

r1

(1− r)[λ′(r)]2dr.

= logr2

(1− r2)2− log

r1

(1− r1)2−

r1

4

r2∫

r1

(1− r)[λ′(r)]2dr.

Thus

logR2(1− r2)2

r2≤ log

R1(1− r1)2

r1−

r1

4

r2∫

r1

(1− r)[λ′(r)]2dr

giving the lemma �


Proof of theorem 3.9.Setψ(r) =(1− r)2

rM(r, f ). Suppose thatθ is so

chosen thatf (r2eiθ) = M(r2, f ) = R2. Then lemma 3 gives

logψ(r2) ≤ log(1− r1)2R1

r1−

r1

4

r2∫

r1

(1− r)[λ′(r)]2 dr.

≤ logψ(r1) −r1

4

r2∫

r1

(1− r)[λ′(r)]2 dr.

sinceR1 ≤ M(r1, f ). 112

Henceψ(r2) ≤ (r1) showing thatψ(r) decreases (weakly) withr. Ifψ(r1) = ψ(r2) thenψ(r) = a constant forr1 ≤ r ≤ r2 and

r2∫

r1

(1− r)[λ′(r)]2dr = 0.

Since 1− r > 0 thereforeλ′(r) = 0 or λ(r) = constant forr1 ≤ r ≤ r2.

Also ψ(r) = constant gives(1− r)2

r| f (reiθ)| = constant= α (say) for

r1 ≤ r ≤ r2. Because from the lemma(1− r)2

r| f (reiθ)| decreases with

increasingr for fixed θ. Therefore ifθ1 is such thatf (r2eiθ1) = M(r2, f )we see that

(1− r2)2

r2M(r2, f ) =

(1− r2)2

r2| f (r2eiθ1)|

≤(1− r1)2

r1| f (r1eiθ1)| ≤

(1− r1)2

r1M(r1, f )

=(1− r2)2

r2M(r2, f ) becauseψ(r) = constant.

Hence(1− r)2

r| f (reiθ)| is constant forr1 ≤ r ≤ r2.

Now

f (reiθ) = Reiλ=

αr

(1− r)2eiλ=

α(reiθ)ei(λ−θ)

[1 − eiθre−iθ]2


Or f (z) =αei(λ−θ)z

(1− ze−iθ)2for z= reiθ, r1 < r < r2.113

Hence by analytic continuation this equation holds in|z| < 1 andsince f (z) ∈ S, αei(λ−θ)

= 1 and sof (z) ≡ f−θ(z) as required.

In all other casesψ(r) decreases strictly with increasingr and

limr→0

ψ(r) = 1 = limM(r, f )

r.

Thusψ(r) < 1, 0 < r < 1 and sinceψ(r) decreasesα = limr→1

ψ(r) exists

andα ≤ ψ

(

12

)

< 1.

Theorem 10. If f (z) ∈ S andα = 0 in theorem 9 then with the notation

of theorem 6 we have forλ >14

an,λ = O(n2λ−1) as n→ ∞ and in

particular if an are the coefficients of f ,an

n→ 0 as n tends to infinity.

Proof. We recall the proof of theorem 6 and the notations of that theo-rem. We proved

(n+ λ)|an,λ| ≤1

n+ λ − 1I1(ρ, ϕ′)

I1(ρ, ϕ′) ≤

λ

2π

π∫

−π

| f ′(ρeiθ)|2| f (ρeiθ)|2t−2

12

λ

2π

π∫

−π

f (ρeiθ)2λ−2tdθ ]12

We can findρ such that 1−1n≤ ρ ≤ 1− 1

2n and

12π

π∫

−π

| f ′(ρeiθ)|2| f (ρeiθ)|2t−2dθ ≤S2t

(

1− 12n , f

)

12

[

(

1− 12n

)2−

(

1− 1n

)2]

1

4t2

By theorem 4114

S2t(r, f ) ≤ 2tM(r, f )2t= O[(1 − r)]−4t


asr → 1. So in the above inequality the right hand side isn(

O[n4t])

=

O[n4t+1] instead ofO(n4t+1).Just as before (Ch. (3.9) of section 3.1)

π∫

−π

| f (ρeiθ)|2λ−2tdθ = O[

n−1+4λ−4t]

and so we now get

n|an,λ| ≤ (n+ λ)|an,λ| = O[

n+12 (1+4t+4t+4λ−1)

]

= O(n2λ) as required.

�

The caseα > 0.Suppose then from now onα > 0 in theorem 9 and we shall develop

a series of asymptotic formulae by a form of the Hardy-Littlewoodmethod, culminating in a formula for thean,λ. We have first

Theorem 11.There existsθ0 in 0 ≤ θ0 ≤ 2π such that(1−r)2| f (reiθ0)| →α as r→ 1.

Proof. Setrn = 1−1n

and chooseθn so thatf (rneiθn) = M(rn, f ).

Then from lemma 3 we have for 0< r < rn

(1− r)2

r| f (reiθn)| ≥

(1− rn)2

rn| f (rneiθn)| = Bn (say)

Let nowθ0 be a limit point of theθn and choose a fixedr in 0 < r < 1. 115

Then(1− r)2

r| f (reiθ0)| = lim

(1− r)2

r| f (reiθn) ≥ lim Bn

asn tends to infinity through a suitable sequence for whichθn → θ0,

since after somen, r < rn and(1− r)2

r| f (reiθn)| ≥ Bn. Also limBn = α

since(1− r)2

rM(r, f )→ α asr → 1. �


Hence| f (reiθ0)| ≥rα

(1− r)2and this is true for 0< r < 1. Thus

limr→1

(1− r)2| f (reiθ0)| ≥ α

On the other handlimr→1

(1− r)2| f (reiθ0)| ≤ lim−−→

(1− r)2M(r, f ) = α.

Thus limr→1

(1− r)2| f (reiθ0)| = α as required.

Theorem 12. If θ0 is as in theorem 11 and f(reiθ0) = R(r)eiλ(r) then

1∫

0

(1− r)[λ′(r)]2dr converges.

Hence if r→ 1 andρ→ 1 while r < ρ <12

(1+ r) we have uniformly

(1− r)2

(1− ρ)2

f (reiθ0)f (ρeiθ0)

→ 1.

Proof. We have by lemma 3 for 0< r1 < r2 < 1

r1

4

r2∫

r1

(1− r)[λ′(r)]2dr ≤ logR1(1− r1)2

r1− log

R2(1− r2)2

r2

Letting r2→ 1 and by theorem 11 we get116

r1

4

1∫

r1

(1− r)[λ′(r)]dr ≤ logR1(1− r1)2

r1α

proving convergence, sinceλ′(r) is bounded in 0≤ r ≤ r1. Now supposethatr is so near 1 that

1∫

r

(1− t)[λ′(t)]2dt ≤ ǫ and r < ρ <12

(r + 1)


Then

|λ(ρ) − λ(r)| =

∣

∣

∣

∣

∣

∣

∣

∣

ρ∫

r

λ′(t)dt

∣

∣

∣

∣

∣

∣

∣

∣

≤

ρ∫

r

(1− t)[λ′(t)]2dt

12

×

ρ∫

r

dt1− t

12

(by Schwarz inequality).

|λ(ρ) − λ(r)| ≤

[

ǫ log1− r1− ρ

]12

≤ (ǫ log 2)12

So |λ(ρ) − λ(r)| → 0 if r, ρ→ 1 being related as in the theorem. Thus

arg(1− r)2

(1− ρ)2


= λ(r) − λ(ρ)→ 0

and∣

∣

∣

∣

∣

∣

(1− r)2

(1− ρ)2


∣

∣

∣

∣

∣

∣

→α

α= 1.

by the previous theorem. The result follows. � 117

Theorem 13. Suppose f(z) ∈ S . [By the hypothesis of theorem 9,α =

limr→1

(1− r)2

rM(r, f ) and]. Supposeα > 0 and θ0 as defined in theorem

11, [θ0 such that0 ≤ θ0 ≤ 2π and (1− r)2| f (reiθ0)| → α as r→ 1] and

ϕ(z) = f (z)λ = zλ∞∑

n=0an,αzn. Then ifλ >

14

nan,λei(λ+n)θ0 ∼

ϕ

[(

1−1n

)

eiθ0

]

Γ(2λ)as n→ ∞.

Assuming the theorem let us first deduce some corollaries.

Corollary 1. We have|an,λ| ∼αλn2λ−1

Γ′(2λ)as n→ ∞whereα ≤ 1. Equality

is possible only in the case when f(z) = fθ(z) defined in theorem 1.


Proof. Taking moduli for the result of the theorem,

n|an,λ | ∼|ϕ

[(

1− 1n

)

eiθ0]

|

Γ(2λ)as n→ ∞

=

| f[(

1− 1n

)

eiθ0]

|λ

Γ(2λ)∼

(αn2)λ

Γ(2λ)

after theorem 11. �

For, the choice ofθ0 is such that (1− r)2 f (reiθ0) → α as r → 1.

Precisely we get

(

1n

)2 ∣

∣

∣

∣

∣

∣

f

[(

1−1n

)

eiθ0

]∣

∣

∣

∣

∣

∣

→ α asn→ ∞.

Hence we get|an,α| ∼n2λ−1

Γ(2λ). This is corollary 1.

Corollary 2. If fk(z) = z+∞∑

n=1akn+1zkn+1 belongs to S , and if k= 1, 2

or 3 then,

|akn+1| ≤Γ(n+ 2/k)Γ(n+ 1)Γ(2/k)

for n ≥ n0.

Equality holds for f(z) =z

(1− zkeiθ)2/kand all n, otherwise strict in-118

equality holds for n> n0( f ).

To prove this note that

f (z) = [ fk(z1/k)]k ∈ S

and

fk(z1/k) = z1/k(

∞∑

0

akn+1zn) = [ f (z)]1/k.

Now either f (z) =z

(1− zeiθ)2and so

=z

(1− zke1)2/k=

∑

Akn+1zkn+1einθ= fk(z)


where

Akn+1 =

2k

(

2k + 1

)

. . . . . .(

2k + n− 1

)

1 · 2 . . . . . .n

Akn+1 =Γ

(

2k + n

)

Γ(2/k)Γ(n+ 1)

∼n

2k−1

Γ(2/k)

Alternatively since in every other caseα < 1 for f (z) so by corollary 1,

|akn+1| ∼α

1k n

2k−1

Γ(2/k)

and sinceα < 1,|akn+1| < Akn+1 for large n.

Note that ifk = 1 we have|an+1| < n+ 1 finally and ifk = 2, |a2n+1| < 1. 119

Let us now go to the proof of theorem 13. We suppose without lossof generality thatθ0 = (0) in theorem 13 for otherwise we can considere−iθ0 f (zeiθ0) instead off (z) andeiλθ0[ f (zeiθ0)]λ instead ofϕ(z).

Setrn =

(

1−1n

)

. Givenǫ > 0 we define a domain∆n = ∆n(ǫ) = {z :

ǫ

n< |1− z| <

1nǫ|arg(1− z)| <

(

π2 − ǫ

)

andαn =f (rn)

n2so that|αn| → α

asn→ ∞ by theorem 11.

Suppose nowfn(z) =αn

(1− z)2


Lemma 4. We have as n→ ∞ uniformly for z∈ ∆n(ǫ), f (z) ∼ fn(z) andf ′(z) ∼ f ′n(z).

Proof. Putz= ln(w) = rn +1n

w, so that (1− z) =1n

(1− w).

Then by this∆n(ǫ) in thez-plane corresponds to∆1(ǫ) in thew plane.Define a functiongn(z) as

gn(w) = (1− w)2 f [ln(w)]f (rn)

Note that for a fixedǫ, ∆n(ǫ) lies in |z| < 1 for largen. Hencegn(w) isdefined in∆1(ǫ) for sufficiently largen. �

Further|gn(w)| = O(1) for w ∈ ∆1(ǫ) andn > n0. In fact if ∆n(ǫ),∆n(1

2ǫ), ∆n(14ǫ) are denoted by∆n, ∆′n andd is the distance from∆′1 to120

the outside of∆′′1 , the distance between∆′n and the exterior of∆′′n isexactlyd/n.

So if ∆′′n lies in |z| < 1. ∆′n lies in |z| < 1 −dn

and∆′′n lies in |z| < 1

for all largen. Now | f (z)| ≤r

(1− r)2; |z| = r and so if|z| < 1−

dn< 1;

1− |z| ≥dn

.

Hence in∆′n| f (z)| = O

(

n2

d2

)

= O(n2) as n → ∞ and f (rn) =

f

(

1−1n

)

∼ αn2 by theorem 11 asn → ∞ and hence it follows that

f [ln(w)]f (rn)

=O(n2)

n2= O(1) asn→ ∞ uniformly for w in ∆′1. i.e., gn(w)

is bounded in∆′1 for largen.

Next choose−1 < w < 0 so thatw is real andrn+1n

w = 1−1n

(1−w) =

ln(w). After theorem 12.

f [ln(w)]f (rn)

∼(1− rn)2

1n2 (1− w)2

=1

(1− w)2as n→ ∞

for the hypothesis of theorem 12 is satisfied because,

|z| =∣

∣

∣

∣

∣

1−1n+

1n

w∣

∣

∣

∣

∣

< rn <12|(z+ 1)|, rn → 1 as n→ ∞


in this manner ensuring the above asymptotic equality.Therefore it follows on bringing (1−w)2 to the left hand sidegn(w) =

(1− w)2 f [ln(w)]f (rn)

tends to one asn→ ∞, for realw satisfying 0> w >

−1.Thus we have obtainedgn(w) = O(1) asw ranges in∆′1 andgn(w)→ 121

1 asn → ∞ on a set ofw having a limit in∆′1. Hence by Vitalis con-vergence theorem (see Titchmarsh p.p. 168)gn(w) → 1, g′n(w) → 0uniformly in∆1 which is contained in∆′1 and satisfies the condition thatit is bounded by a contour that is interior to∆′1.

Now translating back intoz, n2(1− z)2 f (z)f (rn)

tends to 1 asn tends to

infinity for z∈ ∆n(ǫ).i.e. by the definition offn(z)

f (z)fn(z)

→ 1 for z ∈ ∆n(ǫ) as n→ ∞.

Alsod

dw(1− w)2 f [ln(w)]

f (rn)→ 0

sincen(1− z) = (1− w); 1n

ddz

n2(1− z)2 f (z)f (rn)

→ 0

nddz

(1− z)2 f (z)f (rn)

→ 0

ddz

(1− z)2 f (z) = O(n) since f (rn) ∼ αn2 as n→ ∞.

for z in ∆n(ǫ).

Note that in∆n(ǫ)1n

and (1− z) have the same order of magnitude.

Thus, (1− z)2 f ′(z) − 2(1− z) f (z) = 0(n), z∈ ∆n(ǫ) asn→ ∞ 122

f ′(z) =2 f (z)(1− z)

+O(n3), z ∈ ∆n(ǫ) as n→ ∞.

=2 fn(z)(1− z)

+O(n3)


= f ′n(z) +O(n3) = f ′n(z)[1 +O(1)]

because

f ′n(z) =2αn

(1− z)= O(n3).

Therefore, f ′(z) ∼ f ′n(z) uniformly for z ∈ ∆n(ǫ) asn → ∞. Thiscompletes the proof of the lemma.

Lemma 5. With the notation of theorem 4, forλ > 0,

Sλ(r, f ) = αλSλ[r, (1− z)−2] +O(1− r)−2λ as r→ 1

and ifλ >12

Iλ(r, f ) = αλIλ[r, (1− z)−2] +O(1− r)1−2λ as r→ 1

Proof. Let Rn = |αn|n2

ǫ2, ϕn(z) = fn(z)λ = αλn(1 − z)−2λ and∆n(ǫ) as

defined in the previous lemma. Note thatϕn(z) maps∆n(ǫ) onto the123

sectorǫ4λRλn < |w| < Rλn, |arg(w) − λargαn| < (π − 2ǫ)

The area of the image=!∆n(ǫ)

|ϕ′n(z)|2 dx dy

(3.2.1) = λ(π − 2ǫ)R2λn [1 − ǫ8λ]

Also in∆n(ǫ), fn(z) ∼ f (z) and f ′n(z) ∼ f ′(z) and so

ϕ′n(z) = λ f ′n(z)[ fn(z)]λ−1 ∼ λ f ′(z)[ f (z)]λ−1= ϕ′(z)

So we have

(3.2.2)"

∆n(ǫ)

|φ′n(z)|2dx dy∼ λ(π − 2ǫ)R2λn (1− ǫ8λ)

Also since|ϕn(z)| < Rλn in ∆n(ǫ) we have by lemma 4,

|ϕ(z)| < Rλn(1+ ǫ) there for largen.

Now choosen so thatRn−1 < M(r, f ) < Rn. This is possible at least ifris sufficiently near 1, sinceRn→ ∞ with n. �


Let E0 be the set of points of|z| < r outside∆n(ǫ). Then inE0,|ϕ(z)| < Rλn, and in∆n(ǫ), |ϕ(z)| < (1+ ǫ)Rλn. The image of|z| < 1 by f (z)contains no point more than once, and so the length of the image overany circle|w| = ρ is at the most 2πρ.

If we cut the unit circle along the negative real axis, then the imageby f (z)λ covers any circle,|w| = ρ, with length at the mostλ2πρ. Sothe area of the image of the union ofE0 and∆n(ǫ) by ϕ(z) is at mostπλR2λ

n (1+ ǫ)2.That gives, 124"

E0∪∆n(ǫ)

|ϕ′(z)|2dx dy≤ πλR2λn (1+ ǫ)2

Also by (3.2.2), we have for largen"

∆n(ǫ)

|ϕ′(z)|2dx dy> λπ(1− ǫ)(1− ǫ8λ)R2λn

[Note theǫ terms of r.h.s. of this inequality and (3.2.2)]So "

E0

|ϕ′(z)|2dx dy< R2λn πλ

[

(1+ ǫ)2 − (1− ǫ)(1 − ǫ8λ)]

< ǫ1R2λn

whereǫ1 is small if ǫ is small.Similarly, "

E0

|ϕ′n(z)|2dx dy< ǫ′′R2λn

By theorem 4

S2λ(r, f ) =λ2

2π

r∫

0

ρdρ

π∫

−π

| f (ρeiθ)|2λ−2| f ′(ρeiθ)|2dθ

=λ2

2π

"

|z|<r

| f (ρeiθ)|2λ−2| f ′(ρeiθ)|2dx dy


=12π

"

|z|<r

|ϕ′(z)|2dx dy.

Hence125

S2λ(r, f ) − S2λ(r, fn) =2π

"

|z|<r

(

|ϕ′(z)|2 − |ϕ′n(z)|2)

dx dy

=2π

∫

E1

+

∫

E0

whereE1 is the part of|z| < r within ∆n(ǫ). Since inE1ϕ′n(z) ∼ ϕ′(z)

∫

E1

(

|ϕ′(z)|2 − |ϕ′n(z)|2)

dx dy= O∫

E1

|ϕ′n(z)|2dx dy

= O∫

∆n(ǫ)

|ϕ′n(z)|2dx dy

= O(R2n) by (3.2.1)

Also

|

∫

E0

(

|ϕ′(z)|2 − |ϕ′n(z)|2)

dx dy| ≤∫

E0

|ϕ′n(z)|2 dx dy+∫

E0

|ϕ′(z)|2 dx dy

≤ (ǫ′′ + ǫ′)R2n.

Sinceǫ′′ andǫ′ can be made as small as we please by suitable choice ofǫ,

S2λ(r, f ) = S2λ(r, fn) +O(R2λn )

= S2λ(r, fn) +O(R2λn−1) as Rn ∼ Rn−1

= S2λ(r, fn) +OM(r, f )2λ.

i.e. S2λ(r, f ) = S2λ(r, fn) +O(1− r)−4λ asr → 1 (by theorem 3).126


Hence we get

S2λ(r, f ) = |α2λn |S2λ[r, (1− z)−2] +O(1− r)−4λ as r → 1.

which is the first result. Integrating the above from 0 tor, w.r.t. r afterdividing by r, we get

I2λ(r, f ) = |α|2λIλ[r, (1− z)−2] +O(1− r)1−4λ as r → 1, if λ >14.

Hence we get the lemma [replacing 2λ by λ]

Iλ(r, f ) = αλIλ[r, (1− z)−2] +O(1− r)1−2λ as r → 1

if λ >12

.

Lemma 6. If η > 0 and λ >14

we can choose k> 0 so that if

r0 < r < 1, 1−1n≤ r ≤ 1−

12n

∫

k(1−r)≤|θ|≤π

(

| f (reiθ)|2 + | fn(reiθ)|2)

dθ < η(1− r)1−4λ

Let γ andγ ′ be the arcs|θ| ≤ k(1 − r) andk(1 − r) ≤ |θ| ≤ π on|z| = r. If k is any fixed number, we can chooseǫ so small that the arcγ 127

lies in∆n(ε) for 1−1n≤ r ≤ 1−

12n

for largen (see Hayman [1]).

Hence∫

γ

| fn(z)|2λdθ ∼∫

γ

| f (z)|2λdθ

∫

γ

| fn|2λdθ −

∫

γ

| f |2λdθ = O∫

| f |2λdθ

= O{I2λ(r, f )}

= O(1− r)1−4λ


by the application of theorem 4, as 2λ >12

and alsoI2λ(r, f )−I2λ(r, fn) =

O[(1 − r)1−4λ] by the previous lemma.Hence subtraction gives,

(3.2.3)∫

γ ′

| f |2λdθ −∫

γ ′

| fn|2λdθ = O[(1 − r)1−4λ]

Now the integral

∫

γ ′

| fn|2λdθ

π∫

k(1−r)

|αn|2λdθ

(1− 2r cosθ + r2)2λ

≤

π∫

k(1−r)

dθ

(r sinθ/2)4λ

A(λ)

∞∫

k(1−r)

dθ

θ4λ

1(1− 2r cosθ + r2)

≤1

(r sin2 θ/2)2sinθ/2 ≥ Aθ in that range.

∫

γ ′

| fn|2λdθ ≤ A(λ)[k(1− r)]1−4λ(4)

<13η(1− r)1−4λ

by properly choosingk sufficiently large. Hence in virtue of the relation128

(2) we deduce same forf (z) and hence the lemma.We now proceed to prove theorem 13. We have as in lemma 5,

ϕn(z) = fn(z)λ = αλn(1− z)−2λ= αλn

∞∑

0

bmzm

bm =2λ(2λ + 1) . . . . . . (2λ +m− 1)

m!


=Γ(m+ 1+ 2λ)Γ(2λ)Γ(m+ 1)

∼m2λ−1

Γ(2λ)

and as already defined

ϕ(z) = zλ

1+∞∑

m=1

am,λzm

Hence by considering the coefficients ofϕ′(z) andϕ′n(z),

(n+ λ)an,λ − nbnαλn =

12Π

∫

|z|=r

ϕ′(z)zn+λ

−ϕ′n(z)

zn dz

∣

∣

∣

∣

(n+ λ)an,λ − nbnαλn

∣

∣

∣

∣

∣

≤12π

1rn+λ−1

∣

∣

∣

∣

∣

2π∫

0

(

ϕ′(reiθ) − reλiλθϕ′n(reiθ))

dθ

We shall chooser in the range 1−1n≤ r ≤ 1−

12n

as usual. For ak to be

determined, letγ andγ ′ be the arcs|θ| ≤ k(1− r) andk(1− r) ≤ |θ| ≤ πrespectively on|z| = r.

γ, ϕ′ ∼ ϕ′n ◦ (ϕ′n)(rλeiλθ) = O(1− r)−1−2λ as r → 1.

and so 129

ϕ′ − (ϕ′n)(rλeiλθ) = O(1− r)−1−2λ= O(n2λ+1) as r → 1 or as n→∞

The length of the path of integration ofγ = O(1− r) = O

(

1n

)

∫

(

ϕ′(reiθ) − rλeiλθϕ′n(reiθ))

dθ = O(n2λ)

this being true for any fixedk.Again,

|

∫

γ ′

(

ϕ′(reiθ) − rλeiλθϕ′n(reiθ))

dθ|


≤

∫

γ ′

|ϕ′(reiθ)|dθ +∫

γ ′

rλ|ϕ′n(reiθ)|dθ

We now proceed as in theorem 6, chooset such that 2λ − 2t >12

and

using Schwarz’s Inequality.∫

γ ′

|ϕ′(reiθ)|dθ = λ∫

γ ′

| f ′(reiθ)|| f (reiθ)|λ−1dθ

≤ λ

∫

γ

| f ′(reiθ)|2| f (reiθ)|2t−2λdθ

12

∫

γ ′

| f (reiθ)|2λ−2tdθ

12

By the same method as of theorem 6, we can chooser such that the firstintegral isO(n4t+1).

By what we have just seen in lemma 6 the second integral can bemade less thatδη(1− r)1−4λ+4t i.e. (const.)ηn4λ−4t−1 andη can be madeas small as we please. So

∫

|ϕ′(reiθ)|dθ < η′n2λ

using the inequality (8) of theorem 6.130

η′ can be made as small as we please by choosingk large enough∫

γ ′ϕ′n(reiθ)dθ can be dealt with similarly.

So finally we see that

2π∫

0

|[ϕ′(reiθ) − rλeiλθϕ ′n(reiθ)]dθ ≤ [2η ′ +O(1)]n2λ

and sinceη′ can be made as small as we please the right hand side isO(n2λ).


That gives

an,λ =nbnα

λn

n+ λ+O(n2λ−1)

=n

n+ λf (rn)λ

n2λ

n2λ−1

Γ(2λ)+O(n2λ−1)

Therefore,

nan,λ ∼f (rn)λ

Γ(2λ)as required.

This completes the proof of theorem 13. For an extension of theresult to p-valent functions and further results see for example W.K.Hayman [1], [2].

Bibliography

131[1] Ahlfors, L. [1]: Beitrage Zur Theorie der meromorphen Funktio-

nen 7, Congr. Math. Scand. Oslo 1929.

[2] Bazilevic, I.E. [1]: On distortion theorem and coefficients of uni-valent functions. Mat. Sbornik N.S. 28 (70), 283-292 (1951).

[3] Biernacki, M. [1]: Bull. Acad. Polon. Sci. Cl.III 4 (1956) 5-8

[4] Caratheodory, C. [2]: “Theory of Functions of one variable” Vol.ITranslated by Steinhardt

[5] Csillag, P. [1]: Math. Annalen 110 (1935) pp. 745-’52.

[6] Hayman, W.K. [1]: Proc. L.M.S. 1955 p.p. 257-284.

[7] Hayman, W.K. [2]: “Multivalent Functions” Cambridge Univer-sity Press.

[8] Hayman, W.K. and Stewart, F.M. [1]: Proc. Camb. Phil. Soc.1954.

[9] Kaplan, W. [1]: (‘close to convex Schlicht functions’) MichiganMath. J. 1(1953) 169-’85.

[10] Kennedy, P.B. [1]: Proc. L.M.S. Vol. 6 1956.

[11] Little wood, D. [1]: Q.J. of Maths. Vol.9 (1938)

[12] Shimizu, T. [1]: On the Theory of Meromorphic FunctionsJap. J.Maths. 6 (1929).

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